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belka [17]
3 years ago
5

If the temperature of the helium balloon were increased from 30°C to 35°C and the volume of the balloon only expanded from 0.47L

to 1.55L, how much pressure would be inside the balloon?
Chemistry
1 answer:
Andrej [43]3 years ago
6 0

Answer:

R u from ridge?

Explanation:

Mr Wilson trippin

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Hi there, the correct answer to this would be reactivity because color, melting point, and density are all examples of physical properties.

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zavuch27 [327]

<u>Answer:</u> The temperature at which the food will cook is 219.14°C

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=14.7psi\\T_1=273K\\P_2=(14.7+11.8)psi=26.5psi\\T_2=?

Putting values in above equation, we get:

\frac{14.7psi}{273K}=\frac{26.5psi}{T_2}\\\\T_2=\frac{26.5\times 273}{14.7}=492.14K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

492.14=T(^oC)+273\\\\T(^oC)=219.14^oC

Hence, the temperature at which the food will cook is 219.14°C

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It is the crust, the least dense layer.


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