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3 years ago

Can a molecular formula ever be the same as an empirical formula? Give an example.

Chemistry

1 answer:

kirza4 [7]3 years ago

7 0

Answer:

Different compounds can have the same empirical formula. For example, ethylene C 2 H 4 and propylene C 3 H 6 have the same empirical formula, CH 2. In the same manner, compounds can have the same molecular formula. For example, ethanol and dimethyl ether have the same molecular formula, C 2 H 6.

Explanation:

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Katyanochek1 [597]

Answer:

The answer to your question is V2 = 4.97 l

Explanation:

Data

Volume 1 = V1 = 4.40 L Volume 2 =

Temperature 1 = T1 = 19°C Temperature 2 = T2 = 37°C

Pressure 1 = P1 = 783 mmHg Pressure 2 = 735 mmHg

Process

1.- Convert temperature to °K

T1 = 19 + 273 = 292°K

T2 = 37 + 273 = 310°K

2.- Use the combined gas law to solve this problem

P1V1/T1 = P2V2/T2

-Solve for V2

V2 = P1V1T2 / T1P2

-Substitution

V2 = (783 x 4.40 x 310) / (292 x 735)

-Simplification

V2 = 1068012 / 214620

-Result

V2 = 4.97 l

6 0

4 years ago

Energy removal is illustrated in. A. Changing water ice to waterB. Changing water to steamC. Boiling of gasolineD. Evaporation o

Kryger [21]

Answer:

D. Evaporation of sea water

Explanation:

6 0

2 years ago

A sample of an element has a mass of 34.261 grams and a volume of 3.8 cubic centimeters. To which number of significant figures

nikklg [1K]

The answer is (2). To calculate the density, you need to divide the mass with volume. When doing division, the significant figures of result is the minimal of the numbers using before. So the answer is 2.

3 0

4 years ago

During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S

Kay [80]

<u>Answer:</u> The below calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

<u>Explanation:</u>

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of $^{235}UF_6=349.034348g/mol$

Molar mass of $^{238}UF_6=352.041206g/mol$

By taking their ratio, we get:

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}$

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}$

From the above relation, it is clear that rate of effusion of $^{235}UF_6$ is faster than $^{238}UF_6$

Difference in the rate of both the gases, $Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816$

To calculate the percentage increase in the rate, we use the equation:

$\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100$

Putting values in above equation, we get:

$\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%$

The above calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

3 0

3 years ago

what do Mars, Mercury, and Venus have in common? A. They have a gas surface composition and a thick atmosphere. b They have many

mario62 [17]

Answer:

the best guess would have to be C

3 0

3 years ago