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ryzh [129]
3 years ago
6

A sample of Xe takes 75 seconds to effuse out of a container. An unknown gas takes 37 seconds to effuse out of the identical con

tainer under identical conditions. What is the most likely identity of the unknown gas?
a. Br2
b. He
c. O2
d. Kr
Chemistry
2 answers:
Sergio [31]3 years ago
4 0

Answer:

The Answer is (c)

Explanation:

Rashid [163]3 years ago
3 0

Answer:

The molar mass of B is 32.0 g/mol

The unknown gas might be O2 (option C)

Explanation:

Step 1: Data given

A sample of Xe takes 75 seconds to effuse out of a container.

An unknown gas takes 37 seconds to effuse out of the identical container

Molar mass Xe = 131.3 g/mol

Step 2:

Rate A / rate B = √ (MB/MA)

rate = amount / time

The equation includes rate, but you are given times. Since rate is amount/time, we have to invert that side of the equation.

Time B/Time A = √(MB/MA)

⇒with time B = the time the unknown gas needed to effuse  = 37 seconds

⇒with time A = the time needed Xe needed to effuse = 75 seconds

⇒with MB = the molar mass of the unknown gas = TO BE DETERMINED

⇒ with MA = the molar mass of Xe = 131.3

37/75 = √ (MB/131.3)

MB = 32

The molar mass of B is 32.0 g/mol

The unknown gas might be O2

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The number of atoms or molecules in a mole is known as the __________ constant.
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Answer: Avogrado's Constant

Explanation:

One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant. The concept of the mole can be used to convert between mass and number of particles.

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2 years ago
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)

A

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.

B

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.

C

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
Which of the following best defines molar mass?
coldgirl [10]

Answer:

D. the mass of one mole of a substance

Explanation:

The molar mass of a substance is the mass in grams of one mole of the substance.

  • For an element, the molar mass is the relative atoms mass expressed in grams.
  • For example 23g of Na, 40g of oxygen
  • For compounds, molar mass is the gram -formula or gram - molecular weight.
  • This is determined by the addition of its component atomic masses and then expressed in grams.
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