Let's assume that the gas has ideal gas behavior.
Ideal gas law,<span>
PV = nRT
(1)
Where, P is the pressure of the gas (Pa), V is the volume of
the gas (m³), n is the number of moles of gas (mol), R is the
universal gas constant ( 8.314 J mol</span>⁻¹ K⁻<span>¹)
and T is temperature in Kelvin.
</span>n = m/M
(2)
Where, n is number of moles, m is mass and M is
molar mass.
From (1)
and (2),
PV = (m/M) RT
By
rearranging,
M =
(mRT)/PV (3)
P = standard pressure = 1 atm = 101325
pa
V = 0.896 L = 0.896 x 10⁻³ m³
R = 8.314 J mol⁻¹ K⁻¹<span>
T = Standard temperature = 273 K
m = </span>3.87 g = 3.87 x 10⁻³ kg<span>
M = ?
</span><span>
By appying the formula,
M =(</span>3.87 x 10⁻³ kg x 8.314 J mol⁻¹ K⁻¹ x 273 K) /101325 pa x 0.896 x 10⁻³m³
<span>M = 0.0967 kg
M = 96.7 g.
Hence, the molar mass of the gas is 96.7 g.
</span>
Answers:
8.70 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 32.00 44.01
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 18H₂O
m/g: 9.88
(a) Calculate the <em>moles of O₂
</em>
n = 9.88 g O₂ ×1 mol O₂ /32.00 g O₂
n = 0.3088 mol O₂
(b) Calculate the <em>moles of CO₂</em>
The molar ratio is (16 mol CO₂/25 mol O₂)
n = 0.3088 mol O₂ × (16 mol CO₂/25 mol O₂)
n = 0.1976 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
Mass of CO₂ = 0.1976 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
Mass of CO₂ = 8.70 g CO₂
Answer:
The atomic radius corresponds to Sigma
Explanation:
It is the Van Der Waals radius ;)
Answer:2NaF is the correct one. It’s a simple combination and can be be split with relative ease
Explanation:
