A sample of argon gas collected at STP occupies 5.51 L. How many argon atoms are in 1.00 L of this sample?
What is the mass of the mass(g) of 3.01 * 10^22 formula units of Calcium carbonate?
1. How many moles do you have in 1 L? That's 1 L x (1 mol/22.4L) = ? moles.
Then you know that 1 mole of atoms is 6.022E23 atoms. You finish.
2. How many moles are there in 3.01E22 formula units? That's
3.01E22/6.02E23 = ?
Then ? moles x formula unit mass = grams. You finish.
Post your work if you get stuck.
To calculate the number of argon atoms in 1.00 L of the sample, we need to use Avogadro's law and the ideal gas law.
Avogadro's law states that equal volumes of gases at the same temperature and pressure contain the same number of molecules.
The ideal gas law is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
At STP (standard temperature and pressure), the temperature is 273.15 K and the pressure is 1 atm.
First, let's find the number of moles of argon in 5.51 L of the sample using the ideal gas law:
PV = nRT
(1 atm) (5.51 L) = n (0.0821 L·atm/(mol·K)) (273.15 K)
5.51 = n(0.0821)(273.15)
n = 0.206 mol
Now, we can use Avogadro's law to find the number of argon atoms:
1 L / 5.51 L = x atoms / 0.206 mol
x = (1 L / 5.51 L) * (0.206 mol)
x = 0.0373 mol
Finally, we can convert the number of moles to the number of argon atoms using Avogadro's number. Avogadro's number is approximately 6.022 × 10^23 atoms/mol.
Number of argon atoms = (0.0373 mol) * (6.022 × 10^23 atoms/mol)
Number of argon atoms = 2.245 × 10^22 atoms
Therefore, there are approximately 2.245 × 10^22 argon atoms in 1.00 L of the sample.
Now let's calculate the mass of 3.01 * 10^22 formula units of calcium carbonate.
The molar mass of calcium carbonate (CaCO3) is calculated as follows:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (there are three oxygen atoms in calcium carbonate)
Molar mass of CaCO3: (40.08 g/mol) + (12.01 g/mol) + (16.00 g/mol * 3) = 100.09 g/mol
To find the mass of 3.01 * 10^22 formula units of calcium carbonate, we can use the following calculation:
Mass = (3.01 * 10^22) * (100.09 g/mol)
Mass = 3.01309 * 10^24 g
The mass of 3.01 * 10^22 formula units of calcium carbonate is approximately 3.01309 * 10^24 grams.
To find the number of argon atoms in 1.00 L of the sample, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (Standard Pressure at STP = 1 atm)
V = volume (1.00 L)
n = number of moles of the gas
R = ideal gas constant (0.0821 L atm/mol K)
T = temperature (Standard Temperature at STP = 273 K)
First, we need to find the number of moles of argon using the ideal gas law equation:
n = PV / RT
Substituting the given values, we have:
n = (1 atm) * (1.00 L) / (0.0821 L atm/mol K * 273 K) = 0.043 moles
Next, we need to convert moles to atoms. One mole of any substance contains Avogadro's number of particles, which is approximately 6.022 × 10^23 atoms/mol.
Number of atoms = moles * Avogadro's number
Number of atoms = 0.043 moles * 6.022 × 10^23 atoms/mol = 2.599 × 10^22 atoms
Therefore, there are approximately 2.599 × 10^22 argon atoms in 1.00 L of this argon gas sample at STP.
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To calculate the mass of 3.01 × 10^22 formula units of calcium carbonate, we need to know the molar mass of calcium carbonate (CaCO3).
Molar mass of CaCO3 = (molar mass of Ca) + (3 × molar mass of C) + (3 × molar mass of O)
Molar mass of Ca = 40.08 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of CaCO3 = (40.08 g/mol) + (3 × 12.01 g/mol) + (3 × 16.00 g/mol)
= 40.08 g/mol + 36.03 g/mol + 48.00 g/mol
= 124.11 g/mol
Now, we can calculate the mass of 3.01 × 10^22 formula units of CaCO3 using the equation:
Mass = Number of formula units * Molar mass
Mass = (3.01 × 10^22) * (124.11 g/mol)
= 3.73 × 10^24 g
Therefore, the mass of 3.01 × 10^22 formula units of calcium carbonate is approximately 3.73 × 10^24 grams.