1- How many grams of calcium chloride are needed to produce 10.0g of potassium chloride (balance the equation and show work pls) ____CaCl2(aq) + ____K2CO3(aq)-- ____KCl(aq)+____CaCO3(aq) 2- When 10.0g of calcium metal is reacted with water, 5.00g of calcium hydroxide is produced. Balance the fellowing equation and calculate the % yield for the reaction. (SHOW WORK PLS) ____Ca(s) + __H2O(l)---____Ca(OH)2 (aq) +___H2(g) Answered by Jerrie Catlin 6 months ago.

1- How many grams of calcium chloride are needed to produce 10.0g of potassium chloride (balance the equation and show work pls) 1 CaCl2 + 1 K2CO3 --> 2 KCl + 1 CaCO3 use molar mass to find moles: 10.0 g KCl @ 74.551 g/mol = 0.1341 mol KCl find moles of CaCl2: 0.1341 mol KCl @ (1 mol CaCl2) / (2 mol KCl) = 0.06705 moles CaCl2 find grams of CaCl2: 0.06075 moles CaCl2 @ 110.98 g/mol = 7.441 g CaCl2 in a 3 sigfig problem, that rounds off to: 7.44 g CaCl2 ======================================... 2a) Balance the fellowing equation 1 Ca + 2H2O ---> 1 Ca(OH)2 + 1H2 ======================================... Calculate the theoretical yield of Ca(OH)2 1 molar mass of Ca produces 1 molar mass of Ca(OH)2 so 40.078 g Ca --> 74.093 g Ca(OH)2 then 10.0 g Ca @ [74.093 g Ca(OH)2] / [40.078 g Ca] = 18.49 g Ca(OH)2 in a 3 sig fig problem, that rounds off to: 18.5 g Ca(OH)2 ======================================... 2b) if only 5.00g of calcium hydroxide was produced, calculate the % yield for the reaction 5.00 / 18.5 times 100 = 27.0 % yield Answered by Daniella Vanolinda 6 months ago.

you're given the mass of potassium chloride mandatory that's 2.00g a million. stability the equation: 2NiCl3 + 3K2Cl ------> Ni2(CO3)3m+ 6KCl 2. calculate the moles of two.00g KCl first moles KCL= m/MM = 2.00g/74g/mol =0.027 moles of KCl (KCl has a molecular weight (MM)= 74g/mol) 2. calculate the moles of nickel (III) chloride with its meant 2.00g manufactured from KCl from the reaction with K2CO3 (considering that 2NiCl3:6KCl) moles NiCl3= 0.027 moles KCl X2 mole NiCl3/6mole KCl= 9.0^-3moles NiCl3 3. calculate the mass of NiCl3 mass of NiCl3= 9.0^-3moles NiCl3 x 165g/mol= a million.forty 9 g of NiCl3 Answered by Reiko Cordier 6 months ago.

Calcium chloride doesn't come in atoms. It is a molecule. So supposing the question to be: "If you have 4.44 x 10(25) molecules of calcium chloride...". (The alternative would be "If you have 4.44 x 10(25) atoms of calcium and chlorine in the ratio 1:2 ..." which would give an answer one-third that of what follows here.) (4.44 x 10^25 molecules) / (6.023 x 10^23 molecules/mol) x (110.9848 g CaCl2/mol) = 8182 g CaCl2 Answered by Alonzo Oesterle 6 months ago.

The reaction is : CaC2 + 2 H2O >> C2H2 + Ca(OH)2 Molecular weight CaC2 = 64 g/mol 10 g / 64 = 0.156 mole of calcium carbide ( not calcium chloride ) The ratio between CaC2 and C2H2 is 1 : 1so we would get 0.156 mole of C2H2 At STP : T = 273 K and p = 1 atm V = nRT / p = 0.156 x 0.0821 x 273 / 1 = 3.5 L of C2H2 Answered by Boris Meis 6 months ago.

You mean, of course, calcium carbide, CaC2. CaC2 + 2H2O ----> C2H2 + Ca(OH)2 Now you can work out the volume yourself. Answered by Henry Boelke 6 months ago.

10 mM means 10 millimoles = 0.01 mole Calcium Chloride Dihydrate = CaCl2 * 2 H2O One mole of Calcium Chloride Dihydrate contains one mole of CaCl2 and 2 moles of H2O. Let’s determine the mass of CaCl2 in one mole of CaCl2 * 2 H2O. Mass of 2 moles of H2O = 36 g Mass of 1 mole of CaCl2 = 147.02 – 36 = 111.02 Mass of 0.01 mole = 1.1102 grams of CaCl2 Answered by Ezequiel Budney 6 months ago.

Dissolve x gm calcium chloride in 9x gm (9x mL) water where x is chosen according to how much solution you need to make. Answered by Estrella Ruhoff 6 months ago.

.0299 molar answer-->2.ninety 9 moles of chloride/a hundred liters of answer a million.29 molar calcium chloride 696 ml answer ((.0299 mol Cl)/(L Cl answer))*(.696 L answer)*(a million mol CaCl2)/(2 mol Cl)*(L CaCl2 answer)/(a million.29 mol CaCl2))=8.06 ml CaCl2 Answered by Archie Lykke 6 months ago.

Calcium chloride = CaCl2 and CaCl2 ==> Ca^2+ + 2Cl^- 10.4 g CaCl2 x 1 mol/111 g = 0.0937 moles CaCl2 Assuming a density of H2O of 1 g/ml, then... Osmolarity = 0.0937 moles/0.25 L = 0.375 molar = 0.375 x 3 = 1.125 osmolar (3 ions) To calculate freezing and boiling point, first calculate molality (moles/kg) molaliry = 0.0937 moles/0.25 kg = 0.375 molal Since there are 3 ions after dissociation, the van't Hoff factor (i) is 3 Freezing point: ∆T = imK = (3)(0.375)(1.86) = 2.1ºC so new freezing point = -2.1ºC Boiling point: ∆T = imK = (3)(0.375)(0.52) = 0.59ºC so new boiling point = 100.6ºC Answered by Page Marriot 6 months ago.

.217 x 10^27 molecules divided by 6.02 x 10^23 is equal to the number of moles of the substance. Now multiply that by the molecular mass. The formula is CaCl2. Answered by Signe Valenzano 6 months ago.

a 1.23 M calcium chloride releases twice as many chlorides , forming a 2.46 Molar chloride solution using the dilution formula C1V1 = C2V2 (2.46 M)(V1) = (0.0270 M) (308ml) V1 = 3.38ml should be diluted Answered by Fredrick Banke 6 months ago.

moles CaCl2 = 5.549 g / 110.98 g/mol=0.0500 mass water = 10.954 - 5.549=5.405 g moles water = 5.405 g/18.02 g/mol= 0.300 divide by the smallest number 0.0500 / 0.0500 = 1 => CaCl2 0.300 / 0.0500=6 => H2O x = 6 Answered by Shawanna Duemmel 6 months ago.

wait..... what?! Answered by Marline Madaras 6 months ago.

Assuming density of solution to be same as water, 147mL of 10% Calcium Chloride solution would contain 14.7 grams Calcium Chloride. Since molecular weight of Calcium Chloride is 147 grams per mole, therefore 14.7 grams is equivalent to 0.1 mole. Now, coming to Osmoles. In solution, each molecule of Calcium Chloride dissociates to form one calcium ion and two chloride ions, hence producing 3 ions. CaCl2 -> Ca++ and 2 Cl- So, a solution containing 0.1 mole Calcium Chloride will produce 0.3 mole of ions which is same as 0.3 osmoles. 0.3 osmoles = 300 milliosmoles. Answered by Elida Coultrap 6 months ago.

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

209088/001 | CALCIUM CHLORIDE 10% | CALCIUM CHLORIDE | INJECTABLE/INJECTION | 100MG per ML |

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

021117/001 | CALCIUM CHLORIDE 10% IN PLASTIC CONTAINER | CALCIUM CHLORIDE | INJECTABLE/INJECTION | 100MG per ML |

209088/001 | CALCIUM CHLORIDE 10% | CALCIUM CHLORIDE | INJECTABLE/INJECTION | 100MG per ML |

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