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
Answer:therefore the oxidation state of sulphur is +6 in the compound CaSO4.
Explanation:
Answer : The molal freezing point depression constant of liquid X is, 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of liquid X (solvent) = 450 g = 0.450 kg
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X = 
i = Van't Hoff factor = 1 (for non-electrolyte)
= Molal-freezing-point-depression constant = ?
m = molality
Now put all the given values in this formula, we get
Therefore, the molal freezing point depression constant of liquid X is,
I believe it would be D. Electromagnet. It's been a while since I've done this stuff, tho. Hope this helps!!!! :)
Answer:
Percent Yield = 94.237%
Explanation:
CO = Carbon Dioxide = Molar Mass 28g/mol
C = Carbon = 12g/mol
O = Oxygen = 16g/mol
Theoretical yield = 93.7 grams
Actual yield = 88.3 grams
Percent yield =
(actual yield
/theoretical yield
)x100
Percent Yield = (88.3/93.7)x100
Percent Yield = 94.237%
