Let's assume that the gas has ideal gas behavior.
Then we can use ideal gas equation,
PV = nRT
Where, <span>
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the </span></span>gas is,<span>
P = 2.8 atm = 283710 Pa
V = 98 L = 98 x 10</span>⁻³ m³<span>
T = 292 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?
By applying the formula,
283710 Pa x </span>98 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 292 K
<span> n = 11.45 mol
Hence, moles of gas is </span>11.45 mol.
Answer:
0.192 mol.
Explanation:
- To calculate the no. of moles of a substance (n), we use the relation:
<em>n = mass / molar mass.</em>
mass of AsH₃ = 15.0 g.
molar mass of AsH₃ = 77.95 g/mol.
∴ The number of moles in 15.0 g AsH₃ = mass / molar mass = (15.0 g) / (77.95 g/mol) = 0.192 mol.
Answer:
203 grams
Explanation:
<em>It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³).
</em>
<em><u>Using cross multiplication:</u></em>
1.0 mol contains → 6.022 x 10²³ molecules.
??? mol contains → 7.2 x 10²⁴ molecules.
∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.
<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass </em>= (11.96 mol)(17.0 g/mol) = <em>203.3 g ≅ 203.0 g.</em>
To determine the number of ounces of the chocolate cereal that you have eaten, the operation that needs to be performed is simply the multiplication of the initial number of ounce in the box of cereal and the decimal equivalent of the given percentage. If we let n be the answer,
n = (13 oz)(0.45)
Simplifying, this will give us the answer which is equal to 5.85 oz.
<em>Answer: 5.85 oz. </em>
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) =
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).
