<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg
<em>Molar concentration
</em>
Assume you have 1 L solution.
Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)
= 1190 g solution
Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)
= 84.01 g NaHCO3
Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)
= 1.14 mol NaHCO3
<em>c</em> = 1.14 mol/1 L = 1.14 mol/L
<em>Molal concentration</em>
Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg
<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg
Answer is: excess of hydrazine is 16 grams.
Chemical reaction: N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂<span>O(g).
</span>m(N₂H₄) = 80,1 g.
m(N₂O₄) = 92,0 g.
n(N₂H₄) = m(N₂H₄) ÷ M(N₂H₄).
n(N₂H₄) = 80,1 g ÷ 32 g/mol.
n(N₂H₄) = 2,5 mol.
n(N₂O₄) = 92 g ÷ 92 g/mol.
n(N₂O₄) = 1 mol; limiting reactant.
From chemical reaction: n(N₂H₄) : n(N₂O₄) = 2 : 1.
n(N₂H₄) = 2 mol reacts.
Δn(N₂H₄) = 2,5 mol - 2 mol = 0,5 mol.
Δm(N₂H₄) = 0,5 mol · 32 g/mol = 16 g.
Answer : The molar mass of the unknown gas will be 79.7 g/mol
Explanation : To solve this question we can use graham's law;
Now we can use nitrogen as the gas number 2, which travels faster than gas 1;
So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1
We can compare the rates of both the gases;
So here, Rate of gas 2 / Rate of gas 1 = 
Now, 1.687 = square root [
]
When we square both the sides we get;
2.845 = (molar mass 1) / (28.01 g/mol N2)
On rearranging, we get,
2.845 X (28.01 g/mol N2) = Molar mass 1
So the molar mass of unknown gas will be = 79.7 g/mol
Answer:
ans is (2) 2,4- hexadiene
