Thermal conductors and insalators

Chemistry

1 answer:

bija089 [108]2 years ago

6 0

Answer:

are both thermal?

Explanation:

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Consider the following gas phase reaction:

snow_tiger [21]

Answer:

2NO(g) + O2(g) --> 2NO2(g)

now 400 ml of NO × 2 mol of NO2/2 mol of NO

= 400 ml of NO2

now 500 ml of O2 × 2 mol of NO2/1 mol of O2

= 1000 ml of NO2

now 400 ml of NO2 × 1 mol of O2/2 mol of NO

= 200 ml

subtract that from 500 ml of total i.e. 500-200 =300 ml

The total volume of the reaction mixture is 1000 ml -300ml = 700 ml

6 0

2 years ago

What is the MOLAR heat of combustion of methane(CH₄) if 64.00g of methane are burned to heat 75.0 ml of water from 25.00°C to 95

melamori03 [73]

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

$Q = c \times m \times \Delta T$

where,

c: specific heat capacity
m: mass
ΔT: change in the temperature

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

$Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ$

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

$\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol$