Answer to Question #85079 in Chemistry for Kat

Question

2. What is the volume of a sample of laughing gas, N2O, that contains 7.11*1021 molecules of the compound?

3. If a balloon contains 2.14L of neon gas, what mass of neon gas is contained within the balloon?

4. If I hand you 6.92*1025 atoms of tungsten, what is the mass of the sample you receive?

Answer 1

2. The volume of laughing gas, N₂O can be calculated from the equation:

V = V_m × N/N_A,

where V - volume, V_m - molar volume of gas (22.4 l/mol), N - number of molecules, N_A - Avogadro's number (6.02 × 10²³ mo^l-1). So,

V = (22.4L/mol × 7.11 × 10²¹ molecules) / (6.02 × 10²³ mol) = 26.5 × 10^-2 L = 0.265 L.

Answer: 0.265 L.

3. To calculate mass of neon gas we can from the equation:

m=ρ × V,

where m – mass of substance, V – volume of substance, ρ – density of gas (the density of neon gas is 0,9002 g/L). So,

m(Ne)=0,9002 g/L × 2.14 L = 1.93 g

Answer: The mass of neon gas within the balloon is 1.93 g.

4. We can calculate the mass of tungsten from the equation:

m = M × N/N_A,

where m – mass of substance, M - molar mass of matter, N – number of molecules, NA – Avogadro’s number (6.02 × 10²³ mol^-1).

M = Ar = 183.84 g/mol

m=183.84 g/mol × (6.91× 10²⁵ molecules) / (6,02 × 10²³ mol) = 211.05 × 10² g.

Answer: The mass of tungsten is 211.05 ×10² g.