1 mole of any gas occupy 22.4 L at STP (standard temperature and pressure, 0°C and 1 atm).
Let given gases be 1 mole. So their volumes will be the same, 22.4 liters.
Density is the ratio of mass to volume.
By formula; density= mass/volume; d=m/V
To find out masses of gases, do the mole calculation.
By formula; mole= mass/molar mass; n= m/M; m= n*M
Molar masses are calculated as
1. C₂H₆ (ethane) = 2*12 g/mol + 6*1 g/mol= 30 g/mol
2. NO (nitrogen monoxide) = 1*14 g/mol + 1*16 g/mol= 30 g/mol
3. NH₃ (ammonia) = 1*14 g/mol + 3*1 g/mol= 17 g/mol
4. H₂O (water) = 2*1 g/mol + 1*16 g/mol= 18 g/mol
5. SO₂ (sulfur dioxide) = 1*32 g/mol + 2*16 g/mol= 64 g/mol
Use Periodic Table to get atomic mass of elements.
Since their volumes are equal, compounds having the same molar mass will have the same density.
Recall the formula d= m/V.
Ethane and nitrogen monoxide have the same density.
The answer is C₂H₆ and NO.
Answer: 1. 
2. 3 moles of 
: 2 moles of 
3. 0.33 moles of : 0.92 moles of
4. is the limiting reagent and is the excess reagent.
5. Theoretical yield of 
is 29.3 g
Explanation:
To calculate the moles :
The balanced chemical equation is:
According to stoichiometry :
3 moles of require = 2 moles of
Thus 0.33 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 3 moles of give = 2 moles of
Thus 0.33 moles of give = of
Theoretical yield of 
Thus 29.3 g of aluminium chloride is formed.
Answer:
a. 58.5 g/mol
b. 0.1 mol
Explanation:
a.
The molar mass of Na is 23.0 g/mol. The molar mass of Cl is 35.5 g/mol. The molar mass of NaCl is:
M(Na) + M(Cl) = 23.0 g/mol + 35.5 g/mol = 58.5 g/mol
b. A healthy adult should eat no more than 6 g of salt in one day. The moles corresponding to 6 g of NaCl are:
6 g × (1 mol/58.5 g) = 0.1 mol
H₂CO₃ ⇔ HCO₃⁻ + H⁺
I 0.160 0 0
C -x +x +x
E 0.160-x +x +x
Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]
4.3 x 10⁻⁷ = x² / (0.160-x) (x is neglected in 0.160-x = 0.160)
x² = 6.88 x 10⁻⁸
x = 2.62 x 10⁻⁴
HCO₃⁻ ⇔ CO₃⁻² + H⁺
I 2.62 x 10⁻⁴ 0 2.62 x 10⁻⁴
C -x +x +x
E 2.62 x 10⁻⁴ - x +x 2.62 x 10⁻⁴ + x
Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]
5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)
x = 5.6 x 10⁻¹¹
Thus,
[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M
[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M
[CO₃⁻²] = 5.6 x 10⁻¹¹ M
[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M
[OH⁻] = 3.8 x 10⁻¹¹
