1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon.
Hence, in this case the mass of carbon in 8.46 g of CO2:
(12/44) × 8.46 = 2.3073 g
1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
(2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12 = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
Carbon : Hydrogen
0.1923/0.1923 : 0.2889/0.1923
1 : 1.5
(1 : 1.5) 2
= 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3
<span>Answer:
Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either.
Start with your 2 half reactions:
I- --> IO3-
Cr2O72- --> 2 Cr3+
Balance O by adding H2O:
I- + 3 H2O --> IO3-
Cr2O72- --> 2 Cr3+ + 7H2O
Balance H by adding H+:
I- + 3 H2O --> IO3- + 6 H+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O
Balance charge by adding e-:
I- + 3 H2O --> IO3- + 6 H+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O
Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give:
Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
They bond because they want to make their outer electron shells more stable
Hope this helps
Have a happy holidays
