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

016708/001 | SODIUM CHROMATE CR 51 | SODIUM CHROMATE CR-51 | INJECTABLE/INJECTION | 100uCi per ML |

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

013993/001 | CHROMITOPE SODIUM | SODIUM CHROMATE CR-51 | INJECTABLE/INJECTION | 200uCi per ML |

013993/002 | CHROMITOPE SODIUM | SODIUM CHROMATE CR-51 | INJECTABLE/INJECTION | 2mCi per VIAL |

016708/001 | SODIUM CHROMATE CR 51 | SODIUM CHROMATE CR-51 | INJECTABLE/INJECTION | 100uCi per ML |

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By mass, it would be 32.1%. All you have to do is find the molar mass of Cr, and divide that by the total molar mass of Na_2CrO_4. Your result is 51.9961 g/mol (for Cr) divided by 161.973 g/mol (for Na_2CrO_4), which equals .321017, or 32.1017%. NB: The molar mass for Na_2CrO_4 can be found by adding the respective molar masses per quantity; i.e. (molar mass of Na * 2) + (molar mass of Cr) + (molar mass of O * 4) Answered by Jaclyn Kibby 1 year ago.

Na2CrO4 1 mole (162.01g) contains 1 moles (52.01g) of chromium atoms. Hence %Cr = (52.01/162.01) x 100 = 32.10% Answered by Emmie Lagan 1 year ago.

11.2 kg Fe(CrO2) = 11200 g Fe(CrO2). Step 1) Find the molar mass of FeCrO2. Fe = 55.847 g/mol Cr = 51.996 g/mol O = 2(15.9994 g/mol) Total: 139.8418 g/mol Step 2) Find the # of moles of FeCrO2 in 11.2 kg 11200 g FeCrO2 x (1 mole FeCrO2) / (139.8418 g FeCrO2) = 80.09050227 mol FeCrO2 Step 3) Use the mole ratio from the balanced chemical equation to convert moles of FeCrO2 to moles of Na2CrO4. According to the equation: 4 mol FeCrO2 --> 8 mol Na2CrO4 Therefore, the mole ratio is (8 mol Na2CrO4) / (4 mol FeCrO2). 80.09050227 mol FeCrO2 x (8 mol Na2CrO4) / (4 mol FeCrO2) = 160.1810045 mol Na2CrO4 Step 4) Find the molar mass of sodium chromate Na = 2(22.98977 g/mol) = 45.97954 g/mol Cr = 51.996 g/mol O = 4(15.9994 g/mol) = 63.9976 g/mol Total = 161.97314 g/mol Step 5) Convert moles of Na2CrO4 to kilograms of Na2CrO4 160.1810045 mol Na2CrO4 x (161.97314 g) / (1 mol) x (1 kg) / (1000 g) = 25.94502027 kg Sig figs --> 25.9 kg Na2CrO4 B) We already have the number of moles of FeCrO2. We will use this to find the number of moles of O2 needed and its mass. Moles of FeCrO2 = 80.09050227 According to the chemical equation, 4 mol FeCrO2 reacts with 7 mol O2. Therefore, the mole ratio is 7 mol O2 / 4 mol FeCrO2. 80.09050227 mol FeCrO2 x (7 mol O2) / (4 mol FeCrO2) = 140.158379 mol O2. Step 2) Find the molar mass of O2 O = 2(15.9994) = 31.9988 g/mol Step 3) Convert moles of O2 to kilograms of O2 140.158379 mol O2 x (31.9988 g O2) / (1 mol O2) x 1 kg / 1000 g = 4.484899937 kg O2 Sig Figs --> 4.48 kg O2 Answered by Camelia Laumbach 1 year ago.

The basic reaction involved in this process is: 3NaHSO4 + 2(CrO4)-- + 7H+ = Cr2(SO4)3 + 3Na+ + 5H2O Every mole of chromate reduced requires 1.5 (= 3/2) moles of sodium bisulfate. I assume chrome concentrations refer to the mass concentration of Cr6+, not that of the chromate ion (CrO4)--. The atomic mass of Cr is 51 gm/mol, so a 400ppm solution of Cr6+ contains (400*10^-6 gr Cr)/(gm solution)/(51 gm Cr/mole Cr) = 7.84 * 10^-3 mol Cr/(kg solution). If this is a dilute aqueous solution, then we can make the approximation that 1 kg solution = 1 liter solution. Every kg (or liter) of your solution would requre 1.5 * 7.84*10^-3 = 1.176 * 10^-2 moles of NaHSO3 to reduce all the Cr6+ present. The molecular mass of NaHSO3 is 104.06 gm/mole, so you need 1.176*10^-2 * 104.06 = 1.224 gm NaHSO3/kg solution (or per liter of solution if one assumes 1 liter of solution has a mass of 1 kg). Note that the reduction reaction above consumes hydrogen ions, and these must be supplied, otherwise the reaction will not proceed. Typically, sulfuric acid (H2SO4) is used to acidify the solution being treated to a pH of about 2. If the hexavalent chrome concentration in your solution refers to the concentration of chromate ion, then rather than using the atomic mass of Cr (51 gm/mol) to calculate how many moles of Cr are present per kg of solution, one would use the molecular mass of the chromate ion (= 115.99 gm/mol) instead. Answered by Herlinda Nylander 1 year ago.

Insufficient data to do the calculations. You must first give the amount of hex chrome , not it's concentration. Answered by Romona Vandergraph 1 year ago.

write a balanced equation and use stoichiometry to figure it out Answered by Eddie Mobbs 1 year ago.

How many moles of silver perchlorate and how many moles of sodium iodide must be mixed together to produce 5.54 moles of silver iodide? How many moles of lead (ii) nitrate and how many moles of lithium chromate must be mixed together to produce 150 grames of lead (ii) chromate? How many grams of zinc nitrate and how many grams of potassium phosphate must be mixed together to produce 63.28 grams of zinc phosphate? Please help! Answered by Nada Huyard 1 year ago.

1) AgClO4 + NaI ------------> AgI + NaClO4 (you can see that the equation is already balanced) 1 mole of AgClO4 + 1 mole of NaI ---------> 1 mole AgI + 1 mole NaClO4 (I used ratio and proportion) If 1 mole NaClO4 is to 1 mole AgClO4 then 5.54 mole NaClO4 is to 5.54 mole AgClO4 If 1 mole NaClO4 is to 1 mole NaI then 5.54 mole NaClO4 is to 5.54 mole NaI 2) Pb(NO3)2 + Li2CrO4 ----------------> PbCrO4 + 2LiNO3 1 mol of PbCrO4 = Pb = 207.2g Cr = 51.99g O = 15.99 x 4 = 63.96g PbCrO4 = 323.15g 150g PbCrO4 x 1 mol PbCrO4/323.15g = 0.4641807210273866625406158130899 mol PbCrO4 or 0.46 mol (ratio and proportion) Pb(NO3)2 + Li2CrO4 ----------------> PbCrO4 + 2LiNO3 If 1 mol PbCrO4 is to 1 mol Pb(NO3)2 then: 0.46 mol PbCrO4 is to 0.46 mol Pb(NO3)2 If 1 mol PbCrO4 is to 1 mol Li2CrO4 0.46 mol PbCrO4 is to 0.46 mol Li2CrO4 3) Answered by Shan Frutoz 1 year ago.

Please show your work as well: 1. A technician mixes a solution containing silver nitrate with sodium chromate solution. 2.89g of precipitate is formed. What is the mass of silver nitrate in the first solution? 2. A solution containing chromium (III) chloride is analyzed using a reaction with zinc. If 11.5g of zinc react, what mass of chromium (III) chloride was in the first solution? Answered by Leigha Panelli 1 year ago.

First question: First, we need to write a balanced equation for the reaction: 2 AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2 NaNO3(aq) We are told that 2.89 g of precipitate is formed. That precipitate must be Ag2CrO4, since it is only slightly soluble in water, while NaNO3 is soluble because all sodium, and all nitrate compounds are soluble in water. Assuming that AgNO3 is the limiting reactant, we can use dimensional analysis and equation coefficients to convert 2.89 g of Ag2CrO4 to moles of Ag2CrO4, to moles of AgNO3, to grams of AgNO3. We shall need the following equalities to set up conversion factors: 1 mol Ag2CrO4 = 279.7376 g Ag2CrO4 1 mol Ag2CrO4 = 2 mol AgNO3 1 mol AgNO3 = 169.8749 g AgNO3 [(2.89 g AgCrO4)/1][(1 mol Ag2CrO4)/(279.7376 g Ag2CrO4)][(2 mol AgNO3)/(1 mol Ag2CrO4)][(169.8749 AgNO3)/(1 mol AgNO3)] = 3.509989 g AgNO3 or 3.51 g AgNO3 rounded to three significant figures Answer: The mass of silver nitrate in the first solution is about 3.51 grams. Second question: First, we need to write a balanced equation for the reaction: 2 CrCl3(aq) + 3 Zn(s) → 3 ZnCl2(aq) + 2 Cr(s) We are told that 11.5 g of zinc took part in the reaction. Assuming that CrCl3 is the limiting reactant, we can use dimensional analysis and equation coefficients to convert 11.5 g of zinc to moles of zinc, to moles of CrCl3, to grams of CrCl3. We shall need the following equalities to set up conversion factors: 1 mol Zn = 65.37 g Zn 3 mol Zn = 2 mol CrCl3 1 mol CrCl3 = 158.355 g CrCl3 [(11.5 g Zn)/1][(1 mol Zn)/(65.37 g Zn)][(2 mol CrCl3)/(3 mol Zn)][(158.355 g CrCl3)/(1 mol CrCl3)] = 18.572032 g CrCl3 or 18.6 g CrCl3 rounded to three significant figures Answer: The mass of chromium (III) chloride in the first solution was about 18.6 grams. Answered by Rosina Barrington 1 year ago.

1. The reaction is: 2 AgNO3 (aq) + Na2CrO4 (aq) --> 2 NaNO3 (aq) + Ag2CrO4 (s) 2 Ag+ (aq) + CrO4(2-) (aq) --> Ag2CrO4 (s) NOTE: We know that Silver chromate is the percipitate due to solubility rules. Therefore there are 2.89 g of silver chromate. Use this to determine the mass of silver nitrate. ? mass of AgNO3 = 2.89 g Ag2CrO4 (1 mol Ag2CrO4 / 331.71 g Ag2CrO4) (2 mol AgNO3 / 1 mol Ag2CrO4) (169.8631 g AgNO3 / 1 mol AgNO3) = 2.96 g AgNO3 The answer to number one is 2.96 g AgNO3 2. The chemical reaction is: 2 CrCl3 (aq) + 3 Zn (s) --> 3 ZnCl2 (aq) + 2 Cr (s) ? Mass of CrCl3 = 11.5 g Zn (1 mol Zn / 65.409 g Zn) (2 mol CrCl3 / 3 mol Zn) (158.355 g CrCl3 / 1 mol CrCl3) = 18.56 g CrCl3 The answer to number 2 is 18.56 g CrCl3 Answered by Clelia Sprandel 1 year ago.