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Rainbow [258]
3 years ago
7

When 8.5 g of methane (ch4) is burned in a bomb calorimeter (heat capacity = 2.677 × 103 j/°c), the temperature rises from 24.00

to 27.08°c. how much heat is absorbed by the calorimeter?ch4(g) + 2o2(g) → co2(g) + 2h2o(l); ∆h° = –1283.8 kj?
Chemistry
2 answers:
Scrat [10]3 years ago
8 0
The molar mass of methane is 16 g/mol. The heat absorbed by the calorimeter is the sensible heat, which is the heat gained or lost during a temperature change.

Sensible heat = CΔT = (<span>2.677 kJ/°C)(27.08 - 24</span>°C)
<em>Sensible heat = 8.24 kJ</em>
Oliga [24]3 years ago
3 0

Answer is: 8.24 kJ of heat is absorbed by the calorimeter.

ΔT= 27.08°C - 24.00°C.

ΔT = 3.08°C; temperature change of reaction.

m = 8.5 g; mass of methane.

cp= 2.677 kJ/°C, specific heat capacity of the bomb calorimeter.  

Qcal = ΔT(methane) · cp(methane).  

Qcal = 3.08°C · 2.677 kJ/°C.  

Qcal = 8.245 kJ; amount of heat absorbed.

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If we assume a lot of things, like:

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Then we have the relation:

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We know that when P = 0.55 atm, the volume is 5.31 L

Then:

(0.55 atm)*(5.31 L) = constant

Now, when the gas is at standard pressure ( P = 1 atm)

We still have the relation:

P*V = constant = (0.55 atm)*(5.31 L)

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If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

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Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

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w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

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