a) A 750-mL sample of hydrogen exerts a pressure of 822 torr at 325 K. What pressure does it exert if the temperature is raised to 475 K at constant volume?

b) Aluminum oxide, Al2O3, is used as a filler for paints and varnishes as well as in the manufacture of electrical insulators. Calculate the number of moles in 47.51 g of Al2O3.

Answer: 1.20 x 10^3

Explanation: 822/325 = ?/475

Isolate the unknown by multiplying both sides by 475/1 (reciprocal). So you divided 822 by 325 and get 2.529230769. You take that and multiply it by 475 and get 1201, which equals1.20 x 10^3.

1.

(P1V1/T1) = (P2V2/T2)

2.
mols = grams/molar mass

a) To solve this problem, we can use the combined gas law formula:

(P1 * V1) / T1 = (P2 * V2) / T2

Where:
P1 and P2 are the initial and final pressures, respectively,
V1 and V2 are the initial and final volumes, respectively,
T1 and T2 are the initial and final temperatures, respectively.

In this case, the initial pressure P1 is given as 822 torr, the initial volume V1 is 750 mL (which we will convert to liters), and the initial temperature T1 is 325 K. The final temperature T2 is 475 K, and we need to find the final pressure P2.

First, let's convert the initial volume to liters:
V1 = 750 mL = 750 mL / 1000 mL/L = 0.750 L

Next, we can plug the values into the formula:
(822 torr * 0.750 L) / 325 K = (P2 * 0.750 L) / 475 K

Simplifying the equation:
(615 torr L) / 325 K = (P2 * 0.750 L) / 475 K

To find P2, we can cross-multiply:
(615 torr L * 475 K) = (P2 * 0.750 L * 325 K)

Dividing both sides of the equation by (0.750 L * 325 K):
P2 = (615 torr L * 475 K) / (0.750 L * 325 K)

P2 ≈ 694 torr

Therefore, the hydrogen sample will exert a pressure of approximately 694 torr when the temperature is raised to 475 K at constant volume.

b) To calculate the number of moles in a given mass of a compound, we need to use the molar mass of the compound.

The molar mass of Aluminum oxide (Al2O3) can be calculated by summing the atomic masses of all the atoms present in the compound.

Aluminum (Al) atomic mass = 26.98 g/mol
Oxygen (O) atomic mass = 16.00 g/mol

The molar mass of Al2O3 can be calculated as follows:
(2 * Aluminum atomic mass) + (3 * Oxygen atomic mass)
= (2 * 26.98 g/mol) + (3 * 16.00 g/mol)
= 53.96 g/mol + 48.00 g/mol
= 101.96 g/mol

Now, we can use the given mass (47.51 g) and the molar mass (101.96 g/mol) to calculate the number of moles:

Number of moles = Mass / Molar mass
= 47.51 g / 101.96 g/mol

Calculating the number of moles:
Number of moles ≈ 0.466 mol

Therefore, there are approximately 0.466 moles of Al2O3 in 47.51 g of Al2O3.

a) To solve this problem, we will use the combined gas law formula, which relates the pressure, volume, and temperature of a gas.

The combined gas law formula is given as:

(P1 * V1) / T1 = (P2 * V2) / T2

Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure (what we need to find)
V2 = final volume (which is constant in this case)
T2 = final temperature

Let's substitute the given values into the formula:

P1 = 822 torr
V1 = 750 mL = 750 cm³ (since 1 mL = 1 cm³)
T1 = 325 K
T2 = 475 K

(P1 * V1) / T1 = (P2 * V2) / T2

(822 torr * 750 cm³) / 325 K = (P2 * 750 cm³) / 475 K

Now, solve for P2:

(822 torr * 750 cm³ * 475 K) / (325 K * 750 cm³) = P2

P2 ≈ 1200 torr

Therefore, the hydrogen will exert a pressure of approximately 1200 torr when the temperature is raised from 325 K to 475 K at a constant volume.

b) To calculate the number of moles in a given mass of a compound, we need to use the molar mass of the compound.

To calculate the molar mass of aluminum oxide (Al2O3), we need to determine the atomic masses of aluminum (Al) and oxygen (O) from the periodic table.

Aluminum (Al) has an atomic mass of approximately 26.98 g/mol.
Oxygen (O) has an atomic mass of approximately 16.00 g/mol.

The molecular formula of aluminum oxide (Al2O3) indicates that there are 2 aluminum atoms and 3 oxygen atoms.

Now, determine the molar mass (M) of Al2O3:

M = (2 * atomic mass of Al) + (3 * atomic mass of O)
M = (2 * 26.98 g/mol) + (3 * 16.00 g/mol)
M = 53.96 g/mol + 48.00 g/mol
M = 101.96 g/mol

Now we can calculate the number of moles (n) using the formula:

n = mass (in g) / molar mass (in g/mol)

n = 47.51 g / 101.96 g/mol

n ≈ 0.4654 mol

Therefore, there are approximately 0.4654 moles of Al2O3 in 47.51 g of the compound.