The mass of calcium hydroxide that is formed when 10.0 g of CaO reacts with 10.0 g of water is 13.024 grams
calculation
from the equation
CaO + H2O → Ca(OH)2,
1 moles of CaO reacted with 1 moles of H2O to form 1 moles of Ca(OH)2
find the moles of each reactant
moles=mass/molar mass
moles of CaO= 10 g/56 g/mol=0.179 moles
moles of H2O = 10 g/18 g/mol 0.556 moles
CaO is the limiting reagent therefore by use of mole ratio of CaO:Ca(OH)2 which is 1:1 moles of Ca(OH)2 is = 0.179 moles
mass= moles x molar mass
= 0.176 moles x 74 g/mol = 13.024 grams
2 ways to do this
a. find %Cl in CaCl2
2 x 35.45g/mole = 70.9g Cl
70.9g Cl / 110.9g/mole CaCl2 = 63.93% Cl in CaCl2
0.6963 x 145g = 92.7g = mass Cl
b. determine moles CaCl2 present then mass Cl
145g / 110.9g/mole = 1.31moles CaCl2 present
2moles Cl / 1mole CaCl2 x 1.31moles = 2.62moles Cl
2.62moles Cl x 35.45g/mole = 92.7g Cl
Answer: 0.0000332mol
Explanation: 1mole of CCl4 contains 6.02x10^23 molecules.
Therefore, X mol of CCl4 will contain 2 x 10^19 molecules i.e
Xmol of CCl4 = 2 x 10^19/ 6.02x10^23 = 0.0000332mol
I think ita c sense its the same object
Answer:
b. .28 M KCI
Explanation:
Use the dilution formula.
C₁V₁ = C₂V₂
C₂ = C₁V₁/V₂
C₂ = 1.6*0.175 / 1.0
C₂ = 0.28
