The number of moles present in 29.5 grams of argon is 0.74 mole.
The atomic mass of argon is given as;
Ar = 39.95 g/mole
The number of moles present in 29.5 grams of argon is calculated as follows;
39.95 g ------------------------------- 1 mole
29.5 g ------------------------------ ?
Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.
<em>"Your question seems to be missing the correct symbol for the element" </em>
Argon = Ar
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Answer:
N2
Explanation:
We use the ideal gas equation to calculate the number of moles of the diatomic gas. Then from the number of moles we can get
Given:
P = 2atm
1atm = 101,325pa
2atm = 202,650pa
T = 27 degrees Celsius = 27 + 273.15 = 300.15K
V = 2.2L
R = molar gas constant = 8314.46 L.Pa/molK
PV = nRT
Rearranging n = PV/RT
Substituting these values will yield:
n = (202,650 * 2.2)/(8314.46* 300.15)
n = 0.18 moles
To get the molar mass, we simply divide the mass by the number of moles.
5.1/0.18 = 28.5g/mol
This is the closest to the molar mass of diatomic nitrogen N2.
Hence, the gas is nitrogen gas
Question:
<span>A sample of nitrogen gas had a volume of 500mL, a pressure in its closed container of 740 torr and a temperature of 25°c. what was the volume of gas when the temperature was changed to 50°c and the new pressure was 760 torr?
Answer:
Data Given:
V</span>₁ = 500 mL
P₁ = 740 torr
T₁ = 25 °C + 273 = 298 K
V₂ = ?
P₂ = 760 torr
T₂ = 50 °C + 273 = 323 K
Solution:
Let suppose the gas is acting Ideally, then According to Ideal Gas Equation,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = (P₁ V₁ T₂) ÷ (T₁ P₂)
Putting Values,
V₂ = (740 torr × 500 mL × 323 K) ÷ (298 K × 760 torr)
V₂ = 527.68 mL
Answer:
I think A it looks to be the answer
Explanation:
sorry if wrong
