Concrete is formed of sand. Is it incorrect to suggest that some houses are built using concrete blocks?
The masses can be found by substractions:
- Mass of CaSO₄.H2O (hydrate):
16.05 g - 13.56 g = 2.49 g
15.07 g - 13.56 g = 1.51 g
- The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:
2.49 g - 1.51 g = 0.98 g
- The percent of water is found by the formula:
massWater ÷ massHydrate * 100%
0.98 g ÷ 2.49 g * 100% = 39.36%
- The mole of water is calculated using water's molecular weight (18g/mol):
0.98 g ÷ 18 g/mol = 0.054 mol water
- A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)
1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄
- The ratio of mole of water to mole of anhydrate is:
0.054 mol water / 0.011 mol CaSO₄ = 0.49
In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O
The final temperature : 345 K
<h3>
Further explanation
</h3>
Given
475 cm³ initial volume
600 cm³ final volume
Required
The final temperature
Solution
At standard temperature and pressure , T = 273 K and 1 atm
Charles's Law :
When the gas pressure is kept constant, the gas volume is proportional to the temperature
V₁/T₁=V₂/T₂
Input the value :
T₂=(V₂T₁)/V₁
T₂=(600 x 273)/475
T₂=345 K
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Sliding or rolling friction
