MnCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MnCl2 that dissolves.
MnCl2(s) --> Mn+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.92 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.8 M
The answer to this question is [Cl⁻] = 1.8 M
Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
Answer:
37.25 grams/L.
Explanation:
- Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.
<em>M = (no. of moles of KCl)/(volume of the solution (L))</em>
<em></em>
∵ no. of moles of KCl = (mass of KCl)/(molar mass of KCl)
∴ M = [(mass of KCl)/(molar mass of KCl)]/(volume of the solution (L))
∴ (mass of KCl)/(volume of the solution (L)) = (M)*(molar mass of KCl) = (0.5 M)*(74.5 g/mol) = 37.25 g/L.
<em>So, the grams/L of KCl = 37.25 grams/L.</em>
Answer:
60%
Explanation:
M(NH4NO3) = 2*14 +4*1 +3*16 = 80 g/mol
M(3O) = 3*16 = 48 g/mol
(48/80) *100 % =60% oxygen by mass.
