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professor190 [17]
3 years ago
11

Calculate the heat change in kilocalories for condensation of 6.5 kg of steam at 100 ° C

Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
6 0

Answer:

6,500 gm of steam require= 3,510 kilo calories (approx)

Explanation:

Every 1 gram of water at 100° C  absorb 540 calories

So,

Total water = 6.5 kg = 6,500 gram

So,

6,500 gm of steam require = 6,500 x 540

6,500 gm of steam require= 3,510 kilo calories (approx)

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sukhopar [10]

Answer:

\rho _2=0.22g/L

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}

It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L

Regards!

8 0
3 years ago
A sample of gas has a volume of 20.0 liters at 22.0° C. If the pressure remains constant, what is the volume at 100.0° C?
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