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Answer:
54.1 % Ca, 43.2 % O, 2.7% H
Explanation:
Molecular formula for calcium hydroxide is Ca(OH)₂
As we don't have a mass of Ca(OH)₂ to find out the percentage composition, we consider that the question refers to 1 mol of compound.
1 mol of hydroxide weighs 74.08 g
1 mol of hydroxide has 1 mol of Ca, therefore 40.08 g are Ca
2 moles of O therefore 32g are O
2 moles of H therefore 2 g are H
Percentage composition is known as (Mass of element/Total mass) . 100
- (40.08 / 74.08) . 100 = 54.1 %
- (32 / 74.08) . 100 = 43.2 %
- (2 / 74.08) . 100 = 2.7%
Answer:
-205.7kj
Explanation:
Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of the reaction.
The expression for enthalpy for the following reaction will be,
where,
n = number of moles
Now put all the given values in the above expression, we get:
Therefore, the enthalpy of the following reaction is, -205.7kj
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Sandalwood
Explanation:
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