ApplId/ProductId | Drug name | Active ingredient | Form | Strenght |
---|---|---|---|---|

017630/001 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 100uCi |

017630/002 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | SOLUTION/ORAL | 2mCi per ML |

018671/001 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 100uCi |

018671/002 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 200uCi |

018671/003 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 400uCi |

071909/001 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 100uCi |

071910/001 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 200uCi |

ApplId/ProductId | Drug name | Active ingredient | Form | Strenght |
---|---|---|---|---|

017630/001 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 100uCi |

017630/002 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | SOLUTION/ORAL | 2mCi per ML |

018671/001 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 100uCi |

018671/002 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 200uCi |

018671/003 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 400uCi |

071909/001 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 100uCi |

071910/001 | SODIUM IODIDE I 123 | SODIUM IODIDE I-123 | CAPSULE/ORAL | 200uCi |

A licensed doctor will try to answer your question for free as quickly as possible. Free of charge during the beta period.

I mean, like is it sold to normal people that are not scientist or something like that, and is it true that if you mix Potassium Iodide with sugar and burn it, it can explode? These Information Might Help You Get The Answers: Natrium Peroxide: CAS number 1313-60-6 PubChem 14803 EC number 215-209-4 UN number 1504 RTECS number WD3450000 Molecular formula Na2O2 Molar mass 77.98 g/mol Appearance yellow to white powder Density 2.805 g/cm3 Melting point 675 °C Boiling point decomp. Solubility in water reacts violently Crystal structure Hexagonal Thermochemistry Std enthalpy of formation ΔfHo298 −513 kJ/mol Standard molar entropy So29895 J K−1 mol−1 Hazards MSDSExternal MSDS EU Index011-003-00-1 EU classificationOxidant (O) Corrosive (C) R-phrasesR8, R35 S-phrases(S1/2), S8, S27, S39, S45 NFPA 704 021OX Flash pointNon-flammable Related compounds Other cationsLithium peroxide Potassium peroxide Rubidium peroxide Caesium peroxide Related sodium oxidesSodium oxide Sodium superoxide Related compoundsSodium hydroxide Hydrogen peroxide Natrium Peroxide: CAS number 7681-11-0 PubChem4875 RTECS number TT2975000 Molecular formula KI Molar mass 166.0028 g/mol Appearance white crystalline solid Density 3.123 g/cm3 Melting point 681 °C, 954 K, 1258 °F Boiling point 1330 °C, 1603 K, 2426 °F Solubility in water 128 g/100 ml (6 °C) 140 g/100 mL (20 °C) Solubility2 g/100 mL (ethanol) soluble in acetone slightly soluble in ether, ammonia Structure Dipole momenthi Hazards MSDSExternal MSDS EU IndexNot listed NFPA 704 110 Related compounds Other anionsPotassium fluoride Potassium chloride Potassium bromide Other cationsLithium iodide Sodium iodide Rubidium iodide Caesium iodide Answered by Cassie Wienecke 1 year ago.

No, most chemical supply companies will only sell to schools or other institutions, not to individuals. Even chemicals that are not nearly so dangerous as these can be tricky to get. Answered by Clorinda Silago 1 year ago.

$20 is too much for 100 potassium iodide pills, in my humble opinion. But I suppose they are hard to get, due to unrealistic thinking in the US. Of course in Japan they may need them. We won't need them in the US unless we have a nuclear accident here. Some people, though, think that not drinking contaminated milk would be enough precaution even there. If I were in Japan, I'd take the iodine. As far as I'm concerned, it is dishonest to tell people they need this when they don't. Of course it won't hurt them either. But if people, on their own initiative, are willing to pay a high price for them.... I guess you aren't taking advantage of them then. Do you really think you can get $500 for 100 potassium iodide pills? Answered by Jacelyn Handlin 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Johnie Ahrenholtz 1 year ago.

its been 4 months since ive had AP chemistry, but i think this is right. a) calcium acetate, or ca(c2h3O2)2 has a molecular mass of 158.17. molarity= moles of solute/liters of solution. so here, there is .1758 liters of solution. and the molarity is .207. so you set up an equation to find the moles of solute, which will be: .207=(x)/(.1758). you find x=.0364 moles. so once you have found that there are .0364 moles, you need to convert the moles to grams of calcium acetate. the gram formula mass of calcium acetate is 158.17. so you multiply (158.17) by the number of moles (.0364) to get the number of grams you have of calcium acetate. this number is 5.756 grams. so 7.56 grams is the answer. b) here is a similar situation. molarity=(moles of solute)/(liters of solution). there is 500 ml, which is .5 liters. and now to find the moles of potassium iodide you first find its gram formula mass 9or molecular mass) to be 166. then you do 21.1/166 to find you have .1271 moles of potassium iodide. then you simply do (.1271)/(.5) to get .2542 as the molarity of the solution. c)set up an equation like this: .85=(x)/(145.6) just solve for x and find it to be 123.76 and that's your answer. the key for these problems is knowing molarity=moles of solute/liters of solution, and being able to convert from moles to mass using molecular mass. Answered by Jared Silletto 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Janet Brouillet 1 year ago.

(a) .1758 L x .207mole/L = .03639mole Ca(CH3COO)2 weighs 158.1g/mole 158.1g/mole x .03639mole = 5.753g of Ca(CH3COO)2 (b) potassium iodine weighs 166g/mole 21.1 / 166g/mole = .127 mole .127mole / .500L = .254mole/L = 254M (c) 145.6L x .850 mole/L = 123.76moles of NaCN Answered by Timmy Eagen 1 year ago.

molar mass Ca(CH3COO)2 = 158.17 g/mol moles Calcium acetate = 0.1758 L x 0.207 M=0.0364 mass = 0.0364 mol x 158.17 g/mol=5.76 g ( 3 significant figures) molar mass KI = 166.0 g/mol moles KI = 21.1 g/166.0 g/mol= 0.127 M = 0.127 mol/ 0.500 L= 0.254 M ( 3 significant figures) moles NaCN = 145.6 L x 0.850 M=124 ( 3 significant figures) Answered by Leonarda Jay 1 year ago.

Stoichimetry Mass to Mass problems: 1. What mass of sodium chloride is produced when chrloien recats with 0.29 grams of sodium iodide? Balance the equation first. Please show your work, thank you. 2. Determine the mass of carbon dioxide produced when 0.85 grams of butane reacts with oxygen according to the following equation: 2C4H10 + 13O2 -> 8CO2 + 10H20 Thank you! Answered by Bruce Zuanich 1 year ago.

1. We need to look at the balanced equation so we can know the moles of each molecule. Cl and I are diatomic elements, which means they can't exist alone as ions; they need two of each other. Cl2 + NaI -----> I2 + NaCl balanced: Cl2 + 2NaI-----> I2 + 2NaCl Now, we start with the given value, 0.29 grams of NaI. We ALWAYS start with the value they have given us. The other information you will need to know is: Molar Mass NaI: 149.90 g/mole Molar Mass NaCl: 58.43 g/mole 0.29 grams NaI x (1 mole NaI/149.90 grams NaI) x (2 moles NaCl/2 moles NaI) x (1 mole NaI/58.43 g NaI) Your answer should be 3.31 x 10^-5 grams NaCl. 2. Again, we use the molar masses and the coefficients in the balanced equation to solve for this: molar mass C4H10=58.123 g/mole molar mass CO2=44.009 g/mole According to the equation, for every 2 moles of butane (C4H10) you have 8 moles of carbon dioxide (CO2). So we start converting with the value they have given us. 0.85 grams C4H10 x (1 mole C4H10/58.123 g C4H10) x (8 moles CO2/2 moles C4H10) x (44.009 g CO2/1 mole CO2) So your answer should be 2.57 grams of CO2 is formed. Answered by Bennie Bastura 1 year ago.

Use the factor lable method to solve this problem. Start with your given (3.0 g NaHCO3) Multiply by 1 mole NaHCO3/#g in 1 mole of NaHCO3 (you figure this out from periodic table) Multiply this by the molar ratios from your balanced equation with what you want in the numerabor and what you have in the denominator. (1 mol Na2CO3/2 mol NaHCO3) Multiply this by #g Na2CO3/1 mol Na2CO3. Compute your grams using the periodic table. Now. Cancel all like units and the only units you have left should be grams of Na2CO3, which is what you are looking for. Hope this helps. This is a gram-gram problem. Any time you are given grams of a substance, you MUST first convert it to moles as the reaction is a molar/molar relationship. Once you get it to moles, you multiply it by a fraction that is the ratio of the moles of what you want to the moles of what you have from the balanced equations. This gets you problem into terms of moles of what you want. Then all you have to do is convert moles to grams, if your question is asking for grams. If your question asks for moles you do not need to do this last step. CHEMISTRY TEACHER Answered by Amee Abele 1 year ago.

I mean, like is it sold to normal people that are not scientist or something like that, and is it true that if you mix Potassium Iodide with sugar and burn it, it can explode? These Information Might Help You Get The Answers: Natrium Peroxide: CAS number 1313-60-6 PubChem 14803 EC number 215-209-4 UN number 1504 RTECS number WD3450000 Molecular formula Na2O2 Molar mass 77.98 g/mol Appearance yellow to white powder Density 2.805 g/cm3 Melting point 675 °C Boiling point decomp. Solubility in water reacts violently Crystal structure Hexagonal Thermochemistry Std enthalpy of formation ΔfHo298 −513 kJ/mol Standard molar entropy So29895 J K−1 mol−1 Hazards MSDSExternal MSDS EU Index011-003-00-1 EU classificationOxidant (O) Corrosive (C) R-phrasesR8, R35 S-phrases(S1/2), S8, S27, S39, S45 NFPA 704 021OX Flash pointNon-flammable Related compounds Other cationsLithium peroxide Potassium peroxide Rubidium peroxide Caesium peroxide Related sodium oxidesSodium oxide Sodium superoxide Related compoundsSodium hydroxide Hydrogen peroxide Natrium Peroxide: CAS number 7681-11-0 PubChem4875 RTECS number TT2975000 Molecular formula KI Molar mass 166.0028 g/mol Appearance white crystalline solid Density 3.123 g/cm3 Melting point 681 °C, 954 K, 1258 °F Boiling point 1330 °C, 1603 K, 2426 °F Solubility in water 128 g/100 ml (6 °C) 140 g/100 mL (20 °C) Solubility2 g/100 mL (ethanol) soluble in acetone slightly soluble in ether, ammonia Structure Dipole momenthi Hazards MSDSExternal MSDS EU IndexNot listed NFPA 704 110 Related compounds Other anionsPotassium fluoride Potassium chloride Potassium bromide Other cationsLithium iodide Sodium iodide Rubidium iodide Caesium iodide Answered by Krystle Ponsler 1 year ago.

No, most chemical supply companies will only sell to schools or other institutions, not to individuals. Even chemicals that are not nearly so dangerous as these can be tricky to get. Answered by Derrick Saines 1 year ago.

$20 is too much for 100 potassium iodide pills, in my humble opinion. But I suppose they are hard to get, due to unrealistic thinking in the US. Of course in Japan they may need them. We won't need them in the US unless we have a nuclear accident here. Some people, though, think that not drinking contaminated milk would be enough precaution even there. If I were in Japan, I'd take the iodine. As far as I'm concerned, it is dishonest to tell people they need this when they don't. Of course it won't hurt them either. But if people, on their own initiative, are willing to pay a high price for them.... I guess you aren't taking advantage of them then. Do you really think you can get $500 for 100 potassium iodide pills? Answered by Louie Knoblock 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Clementina Cafferty 1 year ago.

its been 4 months since ive had AP chemistry, but i think this is right. a) calcium acetate, or ca(c2h3O2)2 has a molecular mass of 158.17. molarity= moles of solute/liters of solution. so here, there is .1758 liters of solution. and the molarity is .207. so you set up an equation to find the moles of solute, which will be: .207=(x)/(.1758). you find x=.0364 moles. so once you have found that there are .0364 moles, you need to convert the moles to grams of calcium acetate. the gram formula mass of calcium acetate is 158.17. so you multiply (158.17) by the number of moles (.0364) to get the number of grams you have of calcium acetate. this number is 5.756 grams. so 7.56 grams is the answer. b) here is a similar situation. molarity=(moles of solute)/(liters of solution). there is 500 ml, which is .5 liters. and now to find the moles of potassium iodide you first find its gram formula mass 9or molecular mass) to be 166. then you do 21.1/166 to find you have .1271 moles of potassium iodide. then you simply do (.1271)/(.5) to get .2542 as the molarity of the solution. c)set up an equation like this: .85=(x)/(145.6) just solve for x and find it to be 123.76 and that's your answer. the key for these problems is knowing molarity=moles of solute/liters of solution, and being able to convert from moles to mass using molecular mass. Answered by Loria Corbit 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Berry Schuchman 1 year ago.

(a) .1758 L x .207mole/L = .03639mole Ca(CH3COO)2 weighs 158.1g/mole 158.1g/mole x .03639mole = 5.753g of Ca(CH3COO)2 (b) potassium iodine weighs 166g/mole 21.1 / 166g/mole = .127 mole .127mole / .500L = .254mole/L = 254M (c) 145.6L x .850 mole/L = 123.76moles of NaCN Answered by Lavinia Gawron 1 year ago.

molar mass Ca(CH3COO)2 = 158.17 g/mol moles Calcium acetate = 0.1758 L x 0.207 M=0.0364 mass = 0.0364 mol x 158.17 g/mol=5.76 g ( 3 significant figures) molar mass KI = 166.0 g/mol moles KI = 21.1 g/166.0 g/mol= 0.127 M = 0.127 mol/ 0.500 L= 0.254 M ( 3 significant figures) moles NaCN = 145.6 L x 0.850 M=124 ( 3 significant figures) Answered by Novella Mally 1 year ago.

Stoichimetry Mass to Mass problems: 1. What mass of sodium chloride is produced when chrloien recats with 0.29 grams of sodium iodide? Balance the equation first. Please show your work, thank you. 2. Determine the mass of carbon dioxide produced when 0.85 grams of butane reacts with oxygen according to the following equation: 2C4H10 + 13O2 -> 8CO2 + 10H20 Thank you! Answered by Emmanuel Gnas 1 year ago.

1. We need to look at the balanced equation so we can know the moles of each molecule. Cl and I are diatomic elements, which means they can't exist alone as ions; they need two of each other. Cl2 + NaI -----> I2 + NaCl balanced: Cl2 + 2NaI-----> I2 + 2NaCl Now, we start with the given value, 0.29 grams of NaI. We ALWAYS start with the value they have given us. The other information you will need to know is: Molar Mass NaI: 149.90 g/mole Molar Mass NaCl: 58.43 g/mole 0.29 grams NaI x (1 mole NaI/149.90 grams NaI) x (2 moles NaCl/2 moles NaI) x (1 mole NaI/58.43 g NaI) Your answer should be 3.31 x 10^-5 grams NaCl. 2. Again, we use the molar masses and the coefficients in the balanced equation to solve for this: molar mass C4H10=58.123 g/mole molar mass CO2=44.009 g/mole According to the equation, for every 2 moles of butane (C4H10) you have 8 moles of carbon dioxide (CO2). So we start converting with the value they have given us. 0.85 grams C4H10 x (1 mole C4H10/58.123 g C4H10) x (8 moles CO2/2 moles C4H10) x (44.009 g CO2/1 mole CO2) So your answer should be 2.57 grams of CO2 is formed. Answered by Lia Leech 1 year ago.

Use the factor lable method to solve this problem. Start with your given (3.0 g NaHCO3) Multiply by 1 mole NaHCO3/#g in 1 mole of NaHCO3 (you figure this out from periodic table) Multiply this by the molar ratios from your balanced equation with what you want in the numerabor and what you have in the denominator. (1 mol Na2CO3/2 mol NaHCO3) Multiply this by #g Na2CO3/1 mol Na2CO3. Compute your grams using the periodic table. Now. Cancel all like units and the only units you have left should be grams of Na2CO3, which is what you are looking for. Hope this helps. This is a gram-gram problem. Any time you are given grams of a substance, you MUST first convert it to moles as the reaction is a molar/molar relationship. Once you get it to moles, you multiply it by a fraction that is the ratio of the moles of what you want to the moles of what you have from the balanced equations. This gets you problem into terms of moles of what you want. Then all you have to do is convert moles to grams, if your question is asking for grams. If your question asks for moles you do not need to do this last step. CHEMISTRY TEACHER Answered by Kris Colbeck 1 year ago.

I mean, like is it sold to normal people that are not scientist or something like that, and is it true that if you mix Potassium Iodide with sugar and burn it, it can explode? These Information Might Help You Get The Answers: Natrium Peroxide: CAS number 1313-60-6 PubChem 14803 EC number 215-209-4 UN number 1504 RTECS number WD3450000 Molecular formula Na2O2 Molar mass 77.98 g/mol Appearance yellow to white powder Density 2.805 g/cm3 Melting point 675 °C Boiling point decomp. Solubility in water reacts violently Crystal structure Hexagonal Thermochemistry Std enthalpy of formation ΔfHo298 −513 kJ/mol Standard molar entropy So29895 J K−1 mol−1 Hazards MSDSExternal MSDS EU Index011-003-00-1 EU classificationOxidant (O) Corrosive (C) R-phrasesR8, R35 S-phrases(S1/2), S8, S27, S39, S45 NFPA 704 021OX Flash pointNon-flammable Related compounds Other cationsLithium peroxide Potassium peroxide Rubidium peroxide Caesium peroxide Related sodium oxidesSodium oxide Sodium superoxide Related compoundsSodium hydroxide Hydrogen peroxide Natrium Peroxide: CAS number 7681-11-0 PubChem4875 RTECS number TT2975000 Molecular formula KI Molar mass 166.0028 g/mol Appearance white crystalline solid Density 3.123 g/cm3 Melting point 681 °C, 954 K, 1258 °F Boiling point 1330 °C, 1603 K, 2426 °F Solubility in water 128 g/100 ml (6 °C) 140 g/100 mL (20 °C) Solubility2 g/100 mL (ethanol) soluble in acetone slightly soluble in ether, ammonia Structure Dipole momenthi Hazards MSDSExternal MSDS EU IndexNot listed NFPA 704 110 Related compounds Other anionsPotassium fluoride Potassium chloride Potassium bromide Other cationsLithium iodide Sodium iodide Rubidium iodide Caesium iodide Answered by Shenita Beckel 1 year ago.

No, most chemical supply companies will only sell to schools or other institutions, not to individuals. Even chemicals that are not nearly so dangerous as these can be tricky to get. Answered by Theola Pigford 1 year ago.

$20 is too much for 100 potassium iodide pills, in my humble opinion. But I suppose they are hard to get, due to unrealistic thinking in the US. Of course in Japan they may need them. We won't need them in the US unless we have a nuclear accident here. Some people, though, think that not drinking contaminated milk would be enough precaution even there. If I were in Japan, I'd take the iodine. As far as I'm concerned, it is dishonest to tell people they need this when they don't. Of course it won't hurt them either. But if people, on their own initiative, are willing to pay a high price for them.... I guess you aren't taking advantage of them then. Do you really think you can get $500 for 100 potassium iodide pills? Answered by Madeline Apking 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Lisette Hollins 1 year ago.

its been 4 months since ive had AP chemistry, but i think this is right. a) calcium acetate, or ca(c2h3O2)2 has a molecular mass of 158.17. molarity= moles of solute/liters of solution. so here, there is .1758 liters of solution. and the molarity is .207. so you set up an equation to find the moles of solute, which will be: .207=(x)/(.1758). you find x=.0364 moles. so once you have found that there are .0364 moles, you need to convert the moles to grams of calcium acetate. the gram formula mass of calcium acetate is 158.17. so you multiply (158.17) by the number of moles (.0364) to get the number of grams you have of calcium acetate. this number is 5.756 grams. so 7.56 grams is the answer. b) here is a similar situation. molarity=(moles of solute)/(liters of solution). there is 500 ml, which is .5 liters. and now to find the moles of potassium iodide you first find its gram formula mass 9or molecular mass) to be 166. then you do 21.1/166 to find you have .1271 moles of potassium iodide. then you simply do (.1271)/(.5) to get .2542 as the molarity of the solution. c)set up an equation like this: .85=(x)/(145.6) just solve for x and find it to be 123.76 and that's your answer. the key for these problems is knowing molarity=moles of solute/liters of solution, and being able to convert from moles to mass using molecular mass. Answered by Donette Duenas 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Shondra Muccigrosso 1 year ago.

(a) .1758 L x .207mole/L = .03639mole Ca(CH3COO)2 weighs 158.1g/mole 158.1g/mole x .03639mole = 5.753g of Ca(CH3COO)2 (b) potassium iodine weighs 166g/mole 21.1 / 166g/mole = .127 mole .127mole / .500L = .254mole/L = 254M (c) 145.6L x .850 mole/L = 123.76moles of NaCN Answered by Lolita Krajewski 1 year ago.

molar mass Ca(CH3COO)2 = 158.17 g/mol moles Calcium acetate = 0.1758 L x 0.207 M=0.0364 mass = 0.0364 mol x 158.17 g/mol=5.76 g ( 3 significant figures) molar mass KI = 166.0 g/mol moles KI = 21.1 g/166.0 g/mol= 0.127 M = 0.127 mol/ 0.500 L= 0.254 M ( 3 significant figures) moles NaCN = 145.6 L x 0.850 M=124 ( 3 significant figures) Answered by Young Hilovsky 1 year ago.

Stoichimetry Mass to Mass problems: 1. What mass of sodium chloride is produced when chrloien recats with 0.29 grams of sodium iodide? Balance the equation first. Please show your work, thank you. 2. Determine the mass of carbon dioxide produced when 0.85 grams of butane reacts with oxygen according to the following equation: 2C4H10 + 13O2 -> 8CO2 + 10H20 Thank you! Answered by Hank Groman 1 year ago.

1. We need to look at the balanced equation so we can know the moles of each molecule. Cl and I are diatomic elements, which means they can't exist alone as ions; they need two of each other. Cl2 + NaI -----> I2 + NaCl balanced: Cl2 + 2NaI-----> I2 + 2NaCl Now, we start with the given value, 0.29 grams of NaI. We ALWAYS start with the value they have given us. The other information you will need to know is: Molar Mass NaI: 149.90 g/mole Molar Mass NaCl: 58.43 g/mole 0.29 grams NaI x (1 mole NaI/149.90 grams NaI) x (2 moles NaCl/2 moles NaI) x (1 mole NaI/58.43 g NaI) Your answer should be 3.31 x 10^-5 grams NaCl. 2. Again, we use the molar masses and the coefficients in the balanced equation to solve for this: molar mass C4H10=58.123 g/mole molar mass CO2=44.009 g/mole According to the equation, for every 2 moles of butane (C4H10) you have 8 moles of carbon dioxide (CO2). So we start converting with the value they have given us. 0.85 grams C4H10 x (1 mole C4H10/58.123 g C4H10) x (8 moles CO2/2 moles C4H10) x (44.009 g CO2/1 mole CO2) So your answer should be 2.57 grams of CO2 is formed. Answered by Brenda Nofziger 1 year ago.

Use the factor lable method to solve this problem. Start with your given (3.0 g NaHCO3) Multiply by 1 mole NaHCO3/#g in 1 mole of NaHCO3 (you figure this out from periodic table) Multiply this by the molar ratios from your balanced equation with what you want in the numerabor and what you have in the denominator. (1 mol Na2CO3/2 mol NaHCO3) Multiply this by #g Na2CO3/1 mol Na2CO3. Compute your grams using the periodic table. Now. Cancel all like units and the only units you have left should be grams of Na2CO3, which is what you are looking for. Hope this helps. This is a gram-gram problem. Any time you are given grams of a substance, you MUST first convert it to moles as the reaction is a molar/molar relationship. Once you get it to moles, you multiply it by a fraction that is the ratio of the moles of what you want to the moles of what you have from the balanced equations. This gets you problem into terms of moles of what you want. Then all you have to do is convert moles to grams, if your question is asking for grams. If your question asks for moles you do not need to do this last step. CHEMISTRY TEACHER Answered by Bethel Claverie 1 year ago.

I mean, like is it sold to normal people that are not scientist or something like that, and is it true that if you mix Potassium Iodide with sugar and burn it, it can explode? These Information Might Help You Get The Answers: Natrium Peroxide: CAS number 1313-60-6 PubChem 14803 EC number 215-209-4 UN number 1504 RTECS number WD3450000 Molecular formula Na2O2 Molar mass 77.98 g/mol Appearance yellow to white powder Density 2.805 g/cm3 Melting point 675 °C Boiling point decomp. Solubility in water reacts violently Crystal structure Hexagonal Thermochemistry Std enthalpy of formation ΔfHo298 −513 kJ/mol Standard molar entropy So29895 J K−1 mol−1 Hazards MSDSExternal MSDS EU Index011-003-00-1 EU classificationOxidant (O) Corrosive (C) R-phrasesR8, R35 S-phrases(S1/2), S8, S27, S39, S45 NFPA 704 021OX Flash pointNon-flammable Related compounds Other cationsLithium peroxide Potassium peroxide Rubidium peroxide Caesium peroxide Related sodium oxidesSodium oxide Sodium superoxide Related compoundsSodium hydroxide Hydrogen peroxide Natrium Peroxide: CAS number 7681-11-0 PubChem4875 RTECS number TT2975000 Molecular formula KI Molar mass 166.0028 g/mol Appearance white crystalline solid Density 3.123 g/cm3 Melting point 681 °C, 954 K, 1258 °F Boiling point 1330 °C, 1603 K, 2426 °F Solubility in water 128 g/100 ml (6 °C) 140 g/100 mL (20 °C) Solubility2 g/100 mL (ethanol) soluble in acetone slightly soluble in ether, ammonia Structure Dipole momenthi Hazards MSDSExternal MSDS EU IndexNot listed NFPA 704 110 Related compounds Other anionsPotassium fluoride Potassium chloride Potassium bromide Other cationsLithium iodide Sodium iodide Rubidium iodide Caesium iodide Answered by Pansy Zelazo 1 year ago.

No, most chemical supply companies will only sell to schools or other institutions, not to individuals. Even chemicals that are not nearly so dangerous as these can be tricky to get. Answered by Sharon Lorimer 1 year ago.

$20 is too much for 100 potassium iodide pills, in my humble opinion. But I suppose they are hard to get, due to unrealistic thinking in the US. Of course in Japan they may need them. We won't need them in the US unless we have a nuclear accident here. Some people, though, think that not drinking contaminated milk would be enough precaution even there. If I were in Japan, I'd take the iodine. As far as I'm concerned, it is dishonest to tell people they need this when they don't. Of course it won't hurt them either. But if people, on their own initiative, are willing to pay a high price for them.... I guess you aren't taking advantage of them then. Do you really think you can get $500 for 100 potassium iodide pills? Answered by Stephaine Hobby 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Sid Helfrey 1 year ago.

its been 4 months since ive had AP chemistry, but i think this is right. a) calcium acetate, or ca(c2h3O2)2 has a molecular mass of 158.17. molarity= moles of solute/liters of solution. so here, there is .1758 liters of solution. and the molarity is .207. so you set up an equation to find the moles of solute, which will be: .207=(x)/(.1758). you find x=.0364 moles. so once you have found that there are .0364 moles, you need to convert the moles to grams of calcium acetate. the gram formula mass of calcium acetate is 158.17. so you multiply (158.17) by the number of moles (.0364) to get the number of grams you have of calcium acetate. this number is 5.756 grams. so 7.56 grams is the answer. b) here is a similar situation. molarity=(moles of solute)/(liters of solution). there is 500 ml, which is .5 liters. and now to find the moles of potassium iodide you first find its gram formula mass 9or molecular mass) to be 166. then you do 21.1/166 to find you have .1271 moles of potassium iodide. then you simply do (.1271)/(.5) to get .2542 as the molarity of the solution. c)set up an equation like this: .85=(x)/(145.6) just solve for x and find it to be 123.76 and that's your answer. the key for these problems is knowing molarity=moles of solute/liters of solution, and being able to convert from moles to mass using molecular mass. Answered by Dania Kwak 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Danielle Griblin 1 year ago.

(a) .1758 L x .207mole/L = .03639mole Ca(CH3COO)2 weighs 158.1g/mole 158.1g/mole x .03639mole = 5.753g of Ca(CH3COO)2 (b) potassium iodine weighs 166g/mole 21.1 / 166g/mole = .127 mole .127mole / .500L = .254mole/L = 254M (c) 145.6L x .850 mole/L = 123.76moles of NaCN Answered by Kerstin Borkin 1 year ago.

molar mass Ca(CH3COO)2 = 158.17 g/mol moles Calcium acetate = 0.1758 L x 0.207 M=0.0364 mass = 0.0364 mol x 158.17 g/mol=5.76 g ( 3 significant figures) molar mass KI = 166.0 g/mol moles KI = 21.1 g/166.0 g/mol= 0.127 M = 0.127 mol/ 0.500 L= 0.254 M ( 3 significant figures) moles NaCN = 145.6 L x 0.850 M=124 ( 3 significant figures) Answered by Ardis Vinci 1 year ago.

Stoichimetry Mass to Mass problems: 1. What mass of sodium chloride is produced when chrloien recats with 0.29 grams of sodium iodide? Balance the equation first. Please show your work, thank you. 2. Determine the mass of carbon dioxide produced when 0.85 grams of butane reacts with oxygen according to the following equation: 2C4H10 + 13O2 -> 8CO2 + 10H20 Thank you! Answered by Tracy Bartnik 1 year ago.

1. We need to look at the balanced equation so we can know the moles of each molecule. Cl and I are diatomic elements, which means they can't exist alone as ions; they need two of each other. Cl2 + NaI -----> I2 + NaCl balanced: Cl2 + 2NaI-----> I2 + 2NaCl Now, we start with the given value, 0.29 grams of NaI. We ALWAYS start with the value they have given us. The other information you will need to know is: Molar Mass NaI: 149.90 g/mole Molar Mass NaCl: 58.43 g/mole 0.29 grams NaI x (1 mole NaI/149.90 grams NaI) x (2 moles NaCl/2 moles NaI) x (1 mole NaI/58.43 g NaI) Your answer should be 3.31 x 10^-5 grams NaCl. 2. Again, we use the molar masses and the coefficients in the balanced equation to solve for this: molar mass C4H10=58.123 g/mole molar mass CO2=44.009 g/mole According to the equation, for every 2 moles of butane (C4H10) you have 8 moles of carbon dioxide (CO2). So we start converting with the value they have given us. 0.85 grams C4H10 x (1 mole C4H10/58.123 g C4H10) x (8 moles CO2/2 moles C4H10) x (44.009 g CO2/1 mole CO2) So your answer should be 2.57 grams of CO2 is formed. Answered by Christin Perlich 1 year ago.

Use the factor lable method to solve this problem. Start with your given (3.0 g NaHCO3) Multiply by 1 mole NaHCO3/#g in 1 mole of NaHCO3 (you figure this out from periodic table) Multiply this by the molar ratios from your balanced equation with what you want in the numerabor and what you have in the denominator. (1 mol Na2CO3/2 mol NaHCO3) Multiply this by #g Na2CO3/1 mol Na2CO3. Compute your grams using the periodic table. Now. Cancel all like units and the only units you have left should be grams of Na2CO3, which is what you are looking for. Hope this helps. This is a gram-gram problem. Any time you are given grams of a substance, you MUST first convert it to moles as the reaction is a molar/molar relationship. Once you get it to moles, you multiply it by a fraction that is the ratio of the moles of what you want to the moles of what you have from the balanced equations. This gets you problem into terms of moles of what you want. Then all you have to do is convert moles to grams, if your question is asking for grams. If your question asks for moles you do not need to do this last step. CHEMISTRY TEACHER Answered by Roderick Jabaut 1 year ago.

I mean, like is it sold to normal people that are not scientist or something like that, and is it true that if you mix Potassium Iodide with sugar and burn it, it can explode? These Information Might Help You Get The Answers: Natrium Peroxide: CAS number 1313-60-6 PubChem 14803 EC number 215-209-4 UN number 1504 RTECS number WD3450000 Molecular formula Na2O2 Molar mass 77.98 g/mol Appearance yellow to white powder Density 2.805 g/cm3 Melting point 675 °C Boiling point decomp. Solubility in water reacts violently Crystal structure Hexagonal Thermochemistry Std enthalpy of formation ΔfHo298 −513 kJ/mol Standard molar entropy So29895 J K−1 mol−1 Hazards MSDSExternal MSDS EU Index011-003-00-1 EU classificationOxidant (O) Corrosive (C) R-phrasesR8, R35 S-phrases(S1/2), S8, S27, S39, S45 NFPA 704 021OX Flash pointNon-flammable Related compounds Other cationsLithium peroxide Potassium peroxide Rubidium peroxide Caesium peroxide Related sodium oxidesSodium oxide Sodium superoxide Related compoundsSodium hydroxide Hydrogen peroxide Natrium Peroxide: CAS number 7681-11-0 PubChem4875 RTECS number TT2975000 Molecular formula KI Molar mass 166.0028 g/mol Appearance white crystalline solid Density 3.123 g/cm3 Melting point 681 °C, 954 K, 1258 °F Boiling point 1330 °C, 1603 K, 2426 °F Solubility in water 128 g/100 ml (6 °C) 140 g/100 mL (20 °C) Solubility2 g/100 mL (ethanol) soluble in acetone slightly soluble in ether, ammonia Structure Dipole momenthi Hazards MSDSExternal MSDS EU IndexNot listed NFPA 704 110 Related compounds Other anionsPotassium fluoride Potassium chloride Potassium bromide Other cationsLithium iodide Sodium iodide Rubidium iodide Caesium iodide Answered by Jannette Lamphiear 1 year ago.

No, most chemical supply companies will only sell to schools or other institutions, not to individuals. Even chemicals that are not nearly so dangerous as these can be tricky to get. Answered by Arlen Seliski 1 year ago.

$20 is too much for 100 potassium iodide pills, in my humble opinion. But I suppose they are hard to get, due to unrealistic thinking in the US. Of course in Japan they may need them. We won't need them in the US unless we have a nuclear accident here. Some people, though, think that not drinking contaminated milk would be enough precaution even there. If I were in Japan, I'd take the iodine. As far as I'm concerned, it is dishonest to tell people they need this when they don't. Of course it won't hurt them either. But if people, on their own initiative, are willing to pay a high price for them.... I guess you aren't taking advantage of them then. Do you really think you can get $500 for 100 potassium iodide pills? Answered by Shin Ali 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Kenton Pierini 1 year ago.

its been 4 months since ive had AP chemistry, but i think this is right. a) calcium acetate, or ca(c2h3O2)2 has a molecular mass of 158.17. molarity= moles of solute/liters of solution. so here, there is .1758 liters of solution. and the molarity is .207. so you set up an equation to find the moles of solute, which will be: .207=(x)/(.1758). you find x=.0364 moles. so once you have found that there are .0364 moles, you need to convert the moles to grams of calcium acetate. the gram formula mass of calcium acetate is 158.17. so you multiply (158.17) by the number of moles (.0364) to get the number of grams you have of calcium acetate. this number is 5.756 grams. so 7.56 grams is the answer. b) here is a similar situation. molarity=(moles of solute)/(liters of solution). there is 500 ml, which is .5 liters. and now to find the moles of potassium iodide you first find its gram formula mass 9or molecular mass) to be 166. then you do 21.1/166 to find you have .1271 moles of potassium iodide. then you simply do (.1271)/(.5) to get .2542 as the molarity of the solution. c)set up an equation like this: .85=(x)/(145.6) just solve for x and find it to be 123.76 and that's your answer. the key for these problems is knowing molarity=moles of solute/liters of solution, and being able to convert from moles to mass using molecular mass. Answered by Wade Stoffregen 1 year ago.

Hi---Could someone please help me??? I am struggling. PLEASE explain your answers so I can follow you. Thanks to all. All help is greatly appreciated!! :) Calculate each of the following quantities: (a)Grams of solute in 175.8 mL of 0.207 M calcium acetate (b)Molarity of 500. mL of solution containing 21.1 g of potassium iodide (c)Moles of solute in 145.6 L of 0.850 M sodium cyanide Answered by Shery Bohaty 1 year ago.

(a) .1758 L x .207mole/L = .03639mole Ca(CH3COO)2 weighs 158.1g/mole 158.1g/mole x .03639mole = 5.753g of Ca(CH3COO)2 (b) potassium iodine weighs 166g/mole 21.1 / 166g/mole = .127 mole .127mole / .500L = .254mole/L = 254M (c) 145.6L x .850 mole/L = 123.76moles of NaCN Answered by Myrtle Leep 1 year ago.

molar mass Ca(CH3COO)2 = 158.17 g/mol moles Calcium acetate = 0.1758 L x 0.207 M=0.0364 mass = 0.0364 mol x 158.17 g/mol=5.76 g ( 3 significant figures) molar mass KI = 166.0 g/mol moles KI = 21.1 g/166.0 g/mol= 0.127 M = 0.127 mol/ 0.500 L= 0.254 M ( 3 significant figures) moles NaCN = 145.6 L x 0.850 M=124 ( 3 significant figures) Answered by Ambrose Mauffray 1 year ago.

Stoichimetry Mass to Mass problems: 1. What mass of sodium chloride is produced when chrloien recats with 0.29 grams of sodium iodide? Balance the equation first. Please show your work, thank you. 2. Determine the mass of carbon dioxide produced when 0.85 grams of butane reacts with oxygen according to the following equation: 2C4H10 + 13O2 -> 8CO2 + 10H20 Thank you! Answered by Darius Sperger 1 year ago.

1. We need to look at the balanced equation so we can know the moles of each molecule. Cl and I are diatomic elements, which means they can't exist alone as ions; they need two of each other. Cl2 + NaI -----> I2 + NaCl balanced: Cl2 + 2NaI-----> I2 + 2NaCl Now, we start with the given value, 0.29 grams of NaI. We ALWAYS start with the value they have given us. The other information you will need to know is: Molar Mass NaI: 149.90 g/mole Molar Mass NaCl: 58.43 g/mole 0.29 grams NaI x (1 mole NaI/149.90 grams NaI) x (2 moles NaCl/2 moles NaI) x (1 mole NaI/58.43 g NaI) Your answer should be 3.31 x 10^-5 grams NaCl. 2. Again, we use the molar masses and the coefficients in the balanced equation to solve for this: molar mass C4H10=58.123 g/mole molar mass CO2=44.009 g/mole According to the equation, for every 2 moles of butane (C4H10) you have 8 moles of carbon dioxide (CO2). So we start converting with the value they have given us. 0.85 grams C4H10 x (1 mole C4H10/58.123 g C4H10) x (8 moles CO2/2 moles C4H10) x (44.009 g CO2/1 mole CO2) So your answer should be 2.57 grams of CO2 is formed. Answered by Lewis Schumann 1 year ago.

Use the factor lable method to solve this problem. Start with your given (3.0 g NaHCO3) Multiply by 1 mole NaHCO3/#g in 1 mole of NaHCO3 (you figure this out from periodic table) Multiply this by the molar ratios from your balanced equation with what you want in the numerabor and what you have in the denominator. (1 mol Na2CO3/2 mol NaHCO3) Multiply this by #g Na2CO3/1 mol Na2CO3. Compute your grams using the periodic table. Now. Cancel all like units and the only units you have left should be grams of Na2CO3, which is what you are looking for. Hope this helps. This is a gram-gram problem. Any time you are given grams of a substance, you MUST first convert it to moles as the reaction is a molar/molar relationship. Once you get it to moles, you multiply it by a fraction that is the ratio of the moles of what you want to the moles of what you have from the balanced equations. This gets you problem into terms of moles of what you want. Then all you have to do is convert moles to grams, if your question is asking for grams. If your question asks for moles you do not need to do this last step. CHEMISTRY TEACHER Answered by Siobhan Sinagra 1 year ago.