____ occurs when the vapor pressure of a liquid is equal to the external atmospheric pressure.

Chemistry

1 answer:

vaieri [72.5K]2 years ago

8 0

Answer:

huh be more clear

Explanation:

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While finding the number of molecules of oxygen molecules present in 3.65 moles of Na2SO4 the conversion factor used would be

zvonat [6]

Answer

Avogadro's number: One mole of any substance contains 6.022×10²³ molecules

Explanation

While finding the number of moles of oxygen molecules present in 3.65 moles of Na2SO4 the conversion factor used would be Avodagro's number, which is

One mole of any substance contains 6.022×10²³ molecules.

8 0

1 year 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.

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