N=6.98*10²⁴
Nₐ=6.022*10²³ mol⁻¹
n(Mg)=N/Nₐ
m(Mg)=n(Mg)M(Mg)=M(Mg)N/Nₐ
m(Mg)=24.3g/mol*6.98*10²⁴/(6.022*10²³mol⁻¹)=281.7 g
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂
1.01 x 10^24 molecules.
Explanation:
To calculate the number of molecules in a given number of mole, we can simply multiply by Avogadro's number which is equal to 6.022 x 10 ^23.
Therefore,
10 molecules = 1.68 mol x (6.022 x 10^23 molecules) / (1 mol = 1.01 x 10^24) molecules.
I hope this helps :)
Answer:
157.64 L
Explanation:
We'll begin by converting 30 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 30 °C
T(K) = 30 °C + 273
T (K) = 303 K
Next, we shall convert 600 mmHg to atm. This can be obtained as follow:
760 mmHg = 1 atm
Therefore,
600 mmHg = 600 mmHg × 1 atm / 760 mmHg
600 mmHg = 0.789 atm
Finally, we shall determine the volume of the gas. This can be obtained as follow:
Number of mole (n) = 5 moles
Temperature (T) = 303 K
Pressure (P) = 0.789 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
PV = nRT
0.789 × V = 5 × 0.0821 × 303
0.789 × V = 124.3815
Divide both side by 0.789
V = 124.3815 / 0.789
V = 157.64 L
Therefore, the volume of the gas is 157.64 L
At STP condition 1 mol of any ideal gas will have a volume of 22.4L
1.75 mol of F2 x 22.4 L / 1 mol = 39.2 L
