Complete question:
A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate the heat of combustion of methanol in kJ/mol.
Answer:
the heat of combustion of the methanol is 402.31 kJ/mol
Explanation:
Given;
mass of water,
= 100 g
initial temperature of water, t₁ = 28 ⁰C
final temperature of water, t₂ = 58 ⁰C
specific heat capacity of water = 4.184 J/g⁰C
reacting mass of the methanol, m = 1.00 g
molecular mass of methanol = 32.04 g/mol
number of moles = 1 / 32.04
= 0.0312 mol
Apply the principle of conservation of energy;

Therefore, the heat of combustion of the methanol is 402.31 kJ/mol
2+2=4 and I am a pimp. No cap murdergang slat
Answer:
1.373 wt% Ca(OH)₂
Explanation:
Sample mix = 15.0g
Ca(OH)₂(aq) + 2HCl(aq) => CaCl₂(aq) + 2H₂O(l)
moles HCl = 0.2000g / 36 g·mol⁻¹ 0.0056 mol
moles Ca(OH)₂ = 1/2(moles HCl) = 1/2(0.0056 mol) = 0.0028 mol
mass Ca(OH)₂ = 0.0028 mol ( 74 g/mol ) = 0.206 g
mass % Ca(OH)₂ = (0.206/15.0)100% = 1.373 wt%