a) Calculate the number of moles in 17.8 g of the antacid magnesium hydroxide, Mg(OH)2.
b) A 0.850-mole sample of nitrous oxide, a gas used as an anesthetic by dentists, has a volume of 20.46 L at 123°C and 1.35 atm. What would be its volume at 468°C and 1.35 atm?
Mg --> 24.3 g/mol
2 O --> 32
2 H --> 2
sum = 58.3 g/mol
17.8 g * (1 mol/ 58.3 g) = .305 Mol
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123 + 273 = 396 deg K
468 + 273 = 741 deg K
P1 V1/T1 = n R = P2 V2/T2
V2 = V1 (P1/P2) (T2/T1)
V2 = 20.46 (1.35/1.35) (741/396)
a) To calculate the number of moles in a given mass of a substance, we need to use the formula:
moles = mass / molar mass
In this case, we have 17.8 g of magnesium hydroxide (Mg(OH)2).
The molar mass of magnesium (Mg) is 24.305 g/mol.
The molar mass of oxygen (O) is 16.00 g/mol.
The molar mass of hydrogen (H) is 1.008 g/mol.
To calculate the molar mass of magnesium hydroxide, we multiply the molar mass of each element by the number of atoms present in the compound:
(24.305 g/mol + (16.00 g/mol + 1.008 g/mol) * 2) = 58.33 g/mol
Now we can calculate the number of moles:
moles = 17.8 g / 58.33 g/mol ≈ 0.305 moles
Therefore, there are approximately 0.305 moles in 17.8 g of magnesium hydroxide.
b) To calculate the volume at a different temperature (468°C) and pressure (1.35 atm), we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure = 1.35 atm
V1 = initial volume = 20.46 L
T1 = initial temperature in Kelvin = 123°C + 273 = 396 K
P2 = final pressure = 1.35 atm
V2 = final volume (to be calculated)
T2 = final temperature in Kelvin = 468°C + 273 = 741 K
Now we can rearrange the equation to solve for V2:
V2 = (P1 * V1 * T2) / (T1 * P2)
Let's substitute the values and calculate the final volume:
V2 = (1.35 atm * 20.46 L * 741 K) / (396 K * 1.35 atm)
V2 ≈ 34.62 L
Therefore, the volume of the nitrous oxide at 468°C and 1.35 atm would be approximately 34.62 L.
a) To calculate the number of moles, we need to use the formula:
moles = mass (in grams) / molar mass
First, we need to determine the molar mass of magnesium hydroxide, Mg(OH)2.
The molar mass can be calculated by adding up the atomic masses of all the elements in the compound.
Molar mass of Mg = 24.30 g/mol (from the periodic table)
Molar mass of O = 16.00 g/mol (from the periodic table)
Molar mass of H = 1.01 g/mol (from the periodic table)
Molar mass of Mg(OH)2 = (24.30 g/mol x 1) + (16.00 g/mol x 2) + (1.01 g/mol x 2)
= 58.32 g/mol
Now we can calculate the number of moles:
moles = 17.8 g / 58.32 g/mol
≈ 0.305 moles
Therefore, there are approximately 0.305 moles of magnesium hydroxide in 17.8 g.
b) To calculate the volume of nitrous oxide at a different temperature, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
Let's solve for V at the new temperature, knowing that the pressure and number of moles remain constant:
(P1 x V1) / T1 = (P2 x V2) / T2
Plugging in the given values:
P1 = 1.35 atm
V1 = 20.46 L
T1 = 123 + 273 = 396 K
P2 = 1.35 atm
T2 = 468 + 273 = 741 K
To find V2, rearrange the equation to solve for V2:
V2 = (P1 x V1 x T2) / (T1 x P2)
Now substitute the values:
V2 = (1.35 atm x 20.46 L x 741 K) / (396 K x 1.35 atm)
≈ 33.3 L
Therefore, the volume of the nitrous oxide at 468°C and 1.35 atm would be approximately 33.3 L.