All categories

4 years ago

A sample of a gas occupies 2.76 L at 303K. What would the volume be if the temperature was increased to 404K? Assume constant

Chemistry

1 answer:

Artist 52 [7]4 years ago

6 0

Answer:

If the temperature was increased to 404 K, its volume would be 3.68 L.

Explanation:

Charles' Law gives a relationship between the volume and the temperature of the gas at constant temperature. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the temperature.

$V\propto T$

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

Let

$V_1=2.76\ L\\\\T_1=303\ K\\\\T_2=404\ K$

Let $V_2$ is new volume. Using above formula we get :

$V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{2.76\times 404}{303}\\\\V_2=3.68\ L$

If the temperature was increased to 404 K, its volume would be 3.68 L.

You might be interested in

Which of the following is not part of protozoa?

Serga [27]

Answer: A. Cilla Is Correct.

4 0

3 years ago

294 g of potassium dichromate contains 52 g of chromium and 39 g of potassium. What is the mass percent of oxygen in the compoun

IgorLugansk [536]

69.047% is the answer to this question

3 0

3 years ago

Read 2 more answers

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

3 years ago

A chemist mixes sodium with water and witnesses a violent reaction between the metal and water. this is best classified as

Tema [17]

A law is statement about an observed concept. A theory involves the explanation of scientific concepts or principles. A hypothesis is the predicted explanation about some concepts that has to be tested in order to prove it to be right. An observation is the observing the results of a scientific experiment carried out to test an hypothesis.

Here the given statement 'A chemist mixes sodium with water and witnesses a violent reaction between the metal and water,' can be classified as an observation as it explains what the chemists observes as a result of his chemical experiment or test..

8 0

4 years ago

Two moles of an ideal gas are placed in a container whose volume is 2.3 x 10^-3 m3. The absolute pressure of the gas is 6.9 x 10

PtichkaEL [24]

Answer:

$K.E.=1.97\times 10^{-21}\ J$