Answer:
pH = 1.95
Explanation:
For polyprotic acids, it is generally assumed that all H⁺ comes from the 1st ionization step. The amount of H⁺ delivered into solution for the 2nd and 3rd ionization steps are in the order of 10⁻⁴M and 10⁻⁶M respectively and provide very little change in pH from the quantity delivered in the 1st ionization step.
Therefore... the [H⁺] concentraion and pH are computed as follows...
[H⁺] = √Ka₁[H₃AsO₄] = √(2.5 x 10⁻⁴)(0.500) M = 0.1118M
pH = -log[H⁺] = -log(0.01118) = 1.95
I'd say it's A because while the energy cannot be lost, it can not be 100% efficient because if it was we wouldn't have to worry about energy.
In this redox reaction, the Cu goes from oxidation state of (0) to (+2), therefore it oxidises. N in HNO₃ goes from oxidation state of (+5) to N in NO with oxidation state of (+2) and becomes reduced.
Cu acts as the reducing reagent and HNO₃ is the oxidising agent.
oxidation half reaction
Cu ---> Cu²⁺ + 2e --1)
reduction half reaction
4H⁺ + 3e + NO₃⁻ ---> NO + 2H₂O --2)
to balance the number of electrons , 1) x3 and 2) x2
3Cu ---> 3Cu²⁺ + 6e
8H⁺ + 6e + 2NO₃⁻ ---> 2NO + 4H₂O
add the 2 equations
3Cu + 8H⁺ + 2NO₃⁻ ---> 3Cu²⁺ + 2NO + 4H₂O
add 6 nitrate ions to both sides to add up to 8 and form acid with 8H⁺ ions
3Cu + 8HNO₃ ---> 3Cu(NO₃)₂ + 2NO + 4H₂O
Balanced equation for the redox reaction is as follows;
3Cu(s) + 8HNO₃(aq) → 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O<span>(l)
NO has a coefficient of 2
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The percentage of hydrogen in C7H18 is calculated as follows:
[18/(12*7+1*8)]*100=18%
The amount of hydrogen in 5.2moles is given by:(18/100)*5.2=0.94moles
The answer is C. 146g because you add all of the masses of the individual elements and then mulyiply by 1.72 to get your answer.
