What affected the seed of the objects?

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel

n200080 [17]

The question is incomplete, here is the complete question:

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?

A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.

Answer: The mole fraction of gas B (hydrogen gas) is 0.557

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For nitrogen gas:

Given mass of nitrogen gas = 10.25 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen gas}=\frac{10.25g}{28g/mol}=0.366mol$

For hydrogen gas:

Given mass of hydrogen gas = 2.05 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{2.05g}{2g/mol}=1.025mol$

For ammonia gas:

Given mass of ammonia gas = 7.63 g

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia gas}=\frac{7.63g}{17g/mol}=0.449mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B+n_C}$

Moles of gas B (hydrogen gas) = 1.025 moles

Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles

Putting values in above equation, we get:

$\chi_{(H_2)}=\frac{1.025}{1.84}=0.557$

Hence, the mole fraction of gas B (hydrogen gas) is 0.557

6 0

3 years ago

If pea plants with the genotypes RR and rr cross, what genotype will their offspring all have?

Aliun [14]

I think it’ll be Rr

6 0

3 years ago

Read 2 more answers

8. Estimate the maximum volume percent of Methanol vapor that can exist at standard conditions. Vapor pressure = 88.5 mm Hg in a

sashaice [31]

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

It is known that at standard condition, vapor pressure is 760 mm Hg.

And, it is given that methanol vapor pressure in air is 88.5 mm Hg.

Hence, calculate the volume percentage as follows.

Volume percentage = $\frac{\text{given vapor pressure}}{\text{standard vapor pressure}} \times 100$

= $\frac{88.5}{760} \times 100$

= 11.65%

Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.

6 0

3 years ago

Choose the molecular compound among the substances listed below.

agasfer [191]

Answer: Among the listed substances $XeCl_{4}$ is the molecular compound.

Explanation:

A chemical compound formed by the chemical combination of two or more non-metals is called a molecular compound or covalent compound.

For example, Xe and Cl are non-metals. The compound formed by them is $XeCl_{4}$ which is a molecular compound.

A molecular compound is formed by sharing of atoms between the combining atoms.

Whereas NaF, $MnCl_{2}$ and CaO are all ionic compounds as they are formed by chemical combination of a metal and a non-metal.

Thus, we can conclude that among the listed substances $XeCl_{4}$ is the molecular compound.

4 0

3 years ago

Need help please !!!!

Margaret [11]

Answer:

2p

Explanation:

it has 3 dumbell shapes, hence p

you can't determine the principal quantum number by looking at the shape, however bigger or spread orbital means higher value of n

3 0

3 years ago