A sample gas initially occupies 3.35

1. Balance the equation, then answer the following based on the equation

Artemon [7]

Answer:

Q1. a) 4Al + 3O₂ ➟ 2Al₂O₃

b) 7.4 moles

c) 11.1 moles

Explanation:

To balance an equation, ensure that the total number of atoms of each element on both sides are equal.

Al + O₂ ➟ Al₂O₃

On the left side of the arrow, you would find the reactants while the product(s) is found on the left hand side.

Reactants

Al atoms: 1

O atoms: 2

Product

Al atoms: 2

O atoms: 3

After balancing,

4Al + 3O₂ ➟ 2Al₂O₃

We have 4 Al atoms and 6 O atoms on both sides.

b) The balanced equation tells us the mole ratio of Al to Al₂O₃.

Al: Al₂O₃

= 4: 2 (÷2 throughout)

= 2: 1

This means that for every 1 mole of Al₂O₃, 2 moles of Al is needed.

Since we need 3.7 moles of Al₂O₃,

number of moles of Al needed

= 2×3.7

= 7.4

c) 4Al + 3O₂ ➟ 2Al₂O₃

For every 4 moles of Al, 3 moles of O are needed.

For each mole of Al,

number of moles of O needed

= 3÷4

= 0.75

For 14.8 moles of Al,

number of moles of O required

= 0.75 ×14.8

= 11.1

4 0

3 years ago

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

Acetanilide has a solubility of 1.00 g/185 ml in 25 oc water and 1.00 g/20 ml in 100 oc water. what is the minimum volume of wat

Likurg_2 [28]

In the crystallization process the solid compound is dissolved in the solvent at elevated temperature and the crystallize product obtained by slow cooling of the solution. Here the solubility of acetanilide at 100°C is 1g per 20mL of water. Thus to dissolve 500mg of acetanilide at high temperature that is 100°C we need 10mL of water.

Now at 25°C after the re-crystallization there will be some amount of dissolve acetanilide. Which can be calculated as- 185mL of water is needed to dissolve 1g or 1000mg of acetanilide at 25°C. Thus in 10mL of water there will be $\frac{10X1000}{185}=54.05$ gmg of acetanilide.

6 0

3 years ago

How many molecules are in 120 grams of Na2SO4

brilliants [131]

3.06 × 10^23 molecules

5 0

3 years ago

An air mattress is filled with 16.5 moles of air. The air inside the mattress has a temperature of 295 K and a gauge pressure of

postnew [5]

PV=nRT
3.5×X=16.5×0.082×295
X= 114 L
The volume of the air mattress is 114 liters.

7 0

3 years ago

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