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2 years ago

Please answer ASAP!!!

Chemistry

1 answer:

blsea [12.9K]2 years ago

4 0

Answer:

Explanation:

To solve this problem first of all we will understand the specific heat capacity.

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

The substances with higher value of specific heat capacity require more heat to raise the temperature by one degree as compared the substances having low value of specific heat capacity.For example,

The specific heat capacity of gold is 0.031 cal/g.°C and for mercury is 0.033 cal/g.°C. So, mercury take a time to increase its temperature by one degree by absorbing more heat while gold will heat up faster by absorbing less amount of heat.

1) The mercury will change temperature at a much faster rate under the same heating conditions.

This statement is false because specific heat capacity of gold is less than mercury so gold will change the temperature faster.

2) The two metal samples will change temperature at about the same rate.

There is small difference in specific heat of both metals so gold will change the temperature little faster than mercury.

3) The gold would float if placed in the mercury.

The density of gold is greater than mercury so it will sink into mercury.

4) The gold would sink to the bottom if placed in the mercury.

Yes thus statement is true.

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4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K

swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

$Pv =nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 4.3 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

4.3 liter = $4.3\times10^{-3}=4.3\times10^{-3} m^{3}$

And initial temperature of balloon, $T_{1}$ = 350 K

Let the final volume of balloon is $V_{2}$

And as given, final temperature of balloon, $T_{2}$ is 250 K

Now, $V_{1} = KT_{1}$

$4.3\times10^{-3}=K\times350\ (equation\ 1 )$

$V_{2} = KT_{2}$

$=K\times250\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}$

K cancelled by K.

By cross multiplication:

$350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\ = \frac{1075\times10^{-3}}{350} \\ =2.87\times10^{-3}m^{3}$

Now, convert it into liter with the help of calculation done above,

$2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter$

Therefore, volume of the gas in the balloon at 250 K will be 2.87 liter.

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