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3 years ago

A gas that was heated to 150 Celsius has a new volume of 1587.4 L. What was its volume when its temperature was 100 Celsius?

Chemistry

1 answer:

sdas [7]3 years ago

6 0

Answer:

$\boxed {\boxed {\sf 1058.3 \ L}}$

Explanation:

We are asked to find the new volume of a gas after a change in temperature. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The gas was heated to 150 degrees Celsius and had a volume of 1587.4 liters.

$\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{T_2}$

The temperature was 100 degrees Celsius, but the volume is unknown.

$\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C}$

We are solving for the volume at 100 degrees Celsius, so we must isolate the variable V₂. It is being divided by 100°C and the inverse of division is multiplication. Multiply both sides of the equation by 100°C.

$100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C} * 100 \textdegree C$

$100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = V_2$

The units of degrees Celsius cancel.

$100 *\frac {1587.4 \ L }{150 } = V_2$

$100 *10.58266667 \ L = V_2$

$1058.266667 \ L = V_2$

The original measurement of volume has 5 significant figures, so our answer must have the same. For the number we calculated, that is the tenth place. The 6 in the hundredth place to the right tells us to round to 2 up to a 3.

$1058.3 \ L = V_2$

The volume of the gas at 100 degrees Celsius is approximately <u>1058.3 liters.</u>

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weqwewe [10]

Answer:

MAIN IDEA
What the authors meant is not to judge people by their looks.

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The ugly duckling doesn't fit with the other ducklings and grows up, to be a fabulous bird. In the beginning, he is teased because of his looks, in a quote, "As the ugly duckling grew, many animals in the pond criticized him. His feathers were too dark, his neck was too long, and his body was too big. His fellow ducklings refused to recognize that underneath the feathers, the ugly duckling was kind, good, and gentle. The ugly duckling shed many tears because none of the other animals would play with him."

Near the end, however, it states, "He believed that the swans would never be his friends and looked down into the pond in despair. Then, his eyes opened wide, for he saw his reflection. To his great astonishment, the ugly duckling discovered that he was a beautiful, royal swan! The swans welcomed their new friend, whose lovely appearance was matched only by his kind heart. He now felt like he belonged." Throughout the course of the story, the other birds' views change since now they realize how awful they felt for mistreating the ugly duckling.

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THE PRINCESS AND THE PEA

The king and queen didn't believe that she was a princess, so they had her endure the trial of sleeping on twenty mattresses with a pea under the very bottom one. Near the beginning of the story, it states, "The disheveled maiden looked at the king in disbelief. "I am a princess, come to see the prince," she explained, all the while shivering. The king summoned his wife, the queen, for advice. The queen looked at the hideous creature in disbelief, thinking that no princess would ever allow herself to be found in such a state. She thought quickly and came up with a clever plan. The queen put a small pea at the bottom of a bed and piled twenty mattresses on top of it. When the girl arrived at the room, she had to climb a ladder to reach the top of the bed, where she collapsed in exhaustion." This is when the king and queen were in doubt that she was a princess.

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CONCLUSION
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Explanation:

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3 years ago

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Answer:

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Explanation:

Lookup Avogadro's Number: $N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1}$ (three significant figures.)

Lookup the relative atomic mass of $\rm H$ , $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm H$ : $1.008$ .
$\rm S$ : $32.06$ .
$\rm O$ : $15.999$ .

(For example, the relative atomic mass of $\rm H$ is $1.008$ means that the mass of one mole of $\rm H\!$ atoms would be approximately $1.008\!$ grams on average.)

The question counted the number of $\rm H_2SO_4$ molecules without using any unit. Avogadro's Number $N_{\rm A}$ helps convert the unit of that count to moles.

Each mole of $\rm H_2SO_4$ molecules includes exactly $(1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23}$ of these $\rm H_2SO_4 \!$ molecules.

$3.27 \times 10^{23}$ $\rm H_2SO_4$ molecules would correspond to $\displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol$ of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of $\rm H_2SO_4$ gives the mass of each mole of $\rm H_2SO_4\!$ molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

$\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the mass of approximately $0.541389\; \rm mol$ of $\rm H_2SO_4$ :

$\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}$ .

(Rounded to three significant figures.)

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