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nirvana33 [79]
3 years ago
9

How can you remove energy from matter?

Chemistry
1 answer:
valentinak56 [21]3 years ago
7 0

Answer:

you can remove energy from the matter by lowering its temperature

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Muiltplying 2.5 x 10^10 by 3.5 x 10^-7
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<u>Answer: </u>

The value of \left(2.5 \times 10^{10}\right) \times\left(3.5 \times 10^{-7}\right) is 8750

<u>Solution: </u>

\left(2.5 \times 10^{10}\right) \times\left(3.5 \times 10^{-7}\right)

\Rightarrow\left(2.5 \times 10^{10} \times 3.5 \times 10^{-7}\right)

\Rightarrow\left(2.5 \times 3.5 \times\left(10^{10} \times 10^{-7}\right)\right)

\Rightarrow(2.5 \times 3.5) \times\left(10^{10+(-7)}\right)

\Rightarrow(2.5 \times 3.5) \times\left(10^{10+(-7)}\right)

\Rightarrow(2.5 \times 3.5) \times 10^{3}

\Rightarrow 8.75 \times 10^{3}

\Rightarrow 8.75 \times 1000=8750

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3 years ago
A sample of gas in a closed container at a temperature of 76°c and a pressure of 5.0 atm is heated to 399°c. What pressure does
NikAS [45]

<u>Answer:</u> The pressure that the gas exert at high temperature is 9.63 atm

<u>Explanation:</u>

To calculate the final pressure 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=5.0atm\\T_1=76^oC=[76+273]K=349K\\P_2=?\\T_2=399^oC=[273+399]K=672K

Putting values in above equation, we get:

\frac{5.0atm}{349K}=\frac{P_2}{672K}\\\\P_2=9.63atm

Hence, the pressure that the gas exert at high temperature is 9.63 atm

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