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>
V₁ = 400.0 mL
T₁ = 22.0ºC + 273 = 295 K
V₂ = ?
T₂ = 30.0ºC + 273 = 303 K
V₁ / T₁ = V₂ / T₂
400.0 / 295 = V₂ / 303
295 x V₂ = 400.0 x 303
295 V₂ = 121200
V₂ = 121200 / 295
V₂ = 410.84 mL
hope this helps!
Answer:
Option B. 2Mg(s) + O2 (g) —> 2MgO (s)
Explanation:
From the question given above,
We were told that:
2 solid Mg atoms bond with O2 gas to produce solid MgO.
This can be represented by an equation as follow:
2Mg(s) + O2 (g) —> MgO (s)
Next, we shall balance the above equation as follow:
2Mg(s) + O2 (g) —> MgO (s)
There are 2 atoms of Mg on the left side and 1 atom on the right side. It can be balance by putting 2 in front of MgO as shown below:
2Mg(s) + O2 (g) —> 2MgO (s)
Now, the equation is balanced.
Answer is: there are 3.011·10²³ atoms of calcium.
n(Ca) = 0.50 mol; amount of substance(calcium).
Na = 6.022·10²³ 1/mol; Avogadro's constant or number.
N(Ca) = n(Ca) · Na.
N(Ca) = 0.50 mol · 6.022·10²³ 1/mol.
N(Ca) = 3.011·10²³; number of calcium atoms.
The mole is an SI unit which measures the number of particles in substance. One mole is equal to <span><span>6.022</span></span>·<span><span><span>10</span></span></span>²³<span> atoms.</span>
