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3 years ago

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The volume of a 76.52 g sample of methane gas is 32.7 L at 26.9 °C and 4.45 atm. What will its volume be at 95.4 °C and 1.7 atm?

1 answer:

harkovskaia [24]3 years ago

5 0

Answer:

The volume of a 76.52 g sample of methane gas is

32.7 L at 26.9 °C and 4.45 atm. What will its volume be at 95.4 °C and 1.7 atm?

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When a substance undergoes a Chemical Change, can the change be reversed? Explain your answer.

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Answer:When a substance undergoes a chemical change some of the chemical changes can be reversed with a chemical change.

Explanation:

A chemical change changes all of the element but some chemical changes but very few can be changed back to its original substance only with a chemical change.

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Chelsi has talked to her artist friends about how much money they earn each year from working in the arts. She gathers these val

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Answer:

The median would be 6700

Explanation:

Arrange data values from lowest to highest value

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Ordering a data set x1 ≤ x2 ≤ x3 ≤ ... ≤ xn from lowest to highest value, the median x˜ is the data point separating the upper half of the data values from the lower half.

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If n is even the median is the average of the values at positions p and p + 1 where

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If there are 2 data values in the middle the median is the mean of those 2 values.

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What kind of ions do acids release when dissolved in water?

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Virtually all acids release hydrogen ions (or protons) in the water

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For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

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Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

<u>The reaction </u>

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

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Molecular weight: $M_{NO}=30 g/mol$

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$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) <u>Limiting reagent</u>: $NO$

3) <u>Ammount of excess reagent</u>:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

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A physical change _____ change the composition or identity of the substance. will will not 2. An example of a physical change is

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Answer:

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