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, $m_w$ = 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;

$n\Delta H_{methanol} = Q_{water}\\\\n\Delta H_{methanol} = mc\Delta t\\\\n\Delta H_{methanol} = 100 \times 4.184\times (58-28)\\\\n\Delta H_{methanol} = 12,552 \ J\\\\n\Delta H_{methanol} = 12.552 \ kJ\\\\\Delta H_{methanol} = \frac{12.552}{n} \\\\H_{methanol} = \frac{12.552 \ kJ}{0.0312 \ mol} \\\\\Delta H_{methanol} = 402.31 \ kJ/mol$

Therefore, the heat of combustion of the methanol is 402.31 kJ/mol

5 0

3 years ago

Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 32.5°c.

Korvikt [17]

The rms speed of a gas can be calculated using the following rule:
Vrms = sqrt[(3RT) / M] where
R is the gas constant = 8.314 <span>J K^−1 mol^−1
T is the temperature = 32.5 + 273 = 305.5 degree kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens in the equation to get the rms speed as follows:
Vrms = sqrt [(3*8.314*305.5) / 0.028]
Vrms = 521.665 m/sec</span>

5 0

4 years ago

When water reaches the metastable state, the phase of water will be?

Jet001 [13]

Answer:

water, when the metastable state is reached, is cooled below the zero temperature. It freezes abruptly. this is called metastable. They are not at equilibrium per se; as at negative temperatures the only equilibrium state of water is ice.

Explanation:

7 0

3 years ago