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elixir [45]
3 years ago
10

How many kilograms of water at 15oC can be heated to 95oC by burning 143 grams of methane, CH4, assuming that 100% of the heat i

s used to heat the water. The heat of combustion of methane is 891 kJ per mole of methane. (The specific heat of water is 4.184 J/goC.)
Chemistry
1 answer:
mezya [45]3 years ago
8 0

Answer:

23.7 kg is the mass of H₂O that can be heated

Explanation:

First of all, we need to determine the heat used.

891 kJ per mol of methane is released in the combustion, but we don't have 1 mol, we have 143 g

We convert the mass to moles → 143 g . 1mol / 16 g = 8.93 moles

Now, a rule of three:

1 mol of CH₄ release a heat of 891 kJ

8.93 moles of CH₄ will release (8.93 . 891 kJ) / 1 = 7956.63 kJ

If we convert to J → 7956.63 kJ . 1000 J / kJ = 7956630 Joules

Now we determined the heat released, we can apply the calorimetry formula

Q = m . C . (Final T° - Initial T°)

7956630 J = mass . 4.184 J/g°C ( 95°C - 15°C)

7956630 J / 4.184 J/g°C . 80°C = mass

23771 g = mass

If we convert from g to kg → 23771 g . 1kg / 1000 g = 23.7 kg

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.400 moles of CO2 gas are confined in a 5.00-liter container at 25 °C. Calculate the pressure exerted in atmospheres and mm Hg.
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The pressure exerted by 0.400 moles of carbon dioxide in a 5.00 Liter container at 25 °C would be 1.9563 atm or  1486.788 mm Hg.

<h3>The ideal gas law</h3>

According to the ideal gas law, the product of the pressure and volume of a gas is a constant.

This can be mathematically expressed as:

pv = nRT

Where:

p = pressure of the gas

v = volume

n = number of moles

R = Rydberg constant (0.08206 L•atm•mol-1K)

T = temperature.

In this case:

p is what we are looking for.

v = 5.00 L

n = 0.400 moles

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Now, let's make p the subject of the formula of the equation.

p = nRT/v

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Recall that: 1 atm = 760 mm Hg

Thus:

1.9563 atm = 1.9563 x 760 mm Hg

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In other words, the pressure exerted by the gas in atm is 1.9563 atm and in mm HG is 1486.788 mm Hg.

More on the ideal gas law can be found here: brainly.com/question/28257995

#SPJ1

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1 year ago
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