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3 years ago

If the theoretical yield of a reaction is 22.7 g and the actual yield is 21.0 g , what is the percent yield?

1 answer:

4 0

Percentage yield = actual yield/theoretical yield

So divide 21.0 g by 22.7 g and multiply it by 100 to find the percentage yield

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How many moles of NH3 can you make from 6.20 moles of N2?

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3 years ago

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Air pressure is lowest in the lungs during whi h of the following points during inspiration or expiration

AlladinOne [14]

Answer:

Inspiration

Explanation:

This question is on application of Boyle's law; <u>pressure is inversely proportional to volume</u>.when we inhale air, the diaphragm and the muscles in the ribs contract thus increasing the volume in the lungs.Increased volume of the lungs cause the pressure to decrease.During exhaling, the diaphragm and muscles in the ribs relax, making the lungs to recoil and reduce in volume to force air out.Pressure in the lungs is increased than that in the environment making air to move out.

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3 years ago

Write the balance chemical equation for propane.

Mars2501 [29]

It would be 3C + 4H2 -> C3H8

3 0

4 years ago

William measures a test tube and finds that the mass of the test tube is 5g. He places the lone reactant in the test tube and fi

iogann1982 [59]

86 percent is the percent yield for this experiment if he expected to produce 5g of product.

Explanation:

Given that:

mass of test tube = 5 grams

mass of test tube + reactant is 12.5 grams

mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)

mass of reactant = 12.5 -5

= 7.5 grams

when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.

so mass of product formed = 9.3 - 5

= 4. 3 grams of product is formed (actual yield)

However, he expected the product to be 5 grams (theoretical yield)

Percent yield = $\frac{actual yield}{theoretical yield}$ x 100

putting the values in the formula:

percent yield = $\frac{4.3}{5}$ x 100

= 86 %

86 percent is the percent yield.

5 0

3 years ago

c. The reaction Br2 (l) --> Br2 (g) has ΔH = 30.91 kJ/mol and ΔS = 93.3 J/mol·K. Use this information to show (within close a

egoroff_w [7]

Answer:

The answer to your question is given below.

Explanation:

From the question given above, the following data were obtained:

Br₂ (l) —> Br₂(g)

Enthalpy change (ΔH) = 30.91 KJ/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

Next, we shall convert 30.91 KJ/mol to J/mol. This can be obtained as follow:

1 KJ/mol = 1000 J/mol

Therefore,

30.91 KJ/mol = 30.91 × 1000

30.91 KJ/mol = 30910 J/mol

Thus, 30.91 KJ/mol is equivalent to 30910 J/mol.

Finally, we shall determine the boiling temperature of bromine. This can be obtained as follow:

Enthalpy change (ΔH) = 30910 J/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

ΔS = ΔH / T

93.3 = 30910 / T

Cross multiply

93.3 × T = 30910

Divide both side by 93.3

T = 30910 / 93.3

T = 331.29 K

Thus, the boiling temperature of bromine is 331.29 K

6 0

3 years ago