Answer:
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Explanation:
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MAl₂(SO₄)₃·xH₂O:
(mAl×2) + (mS×3) + (mO×12) + (mH₂O×x)
(27×2)+(32×3)+(16×12)+(x×18) = 342 + 18x [g]
mAl₂: 27×2 = 54 [g]
54g ---------- 13,63%
342+18x ---- 100%
0,1363(342+18x) = 54
46,6146 + 2,4534x = 54
2,4534x = 7,3854
x ≈ 3
>>> Al₂(SO₄)₃·3H₂O <<<<
:)
Answer:
the Molar heat of Combustion of diphenylacetylene
= 
Explanation:
Given that:
mass of diphenylacetylene
= 0.5297 g
Molar Mass of diphenylacetylene
= 178.21 g/mol
Then number of moles of diphenylacetylene
= 
= 
= 0.002972 mol
By applying the law of calorimeter;
Heat liberated by 0.002972 mole of diphenylacetylene
= Heat absorbed by
+ Heat absorbed by the calorimeter
Heat liberated by 0.002972 mole of diphenylacetylene
= msΔT + cΔT
= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C
= 17756.48 J + 2842.39 J
= 20598.87 J
Heat liberated by 0.002972 mole of diphenylacetylene
= 20598.87 J
Heat liberated by 1 mole of diphenylacetylene
will be = 
= 6930979.139 J/mol
= 6930.98 kJ/mol
Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene
= 