As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane
Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ
X = 0.970 moles of Ocatne
So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 0.970 mol × 114.23 g/mol
Mass = 110.83 g of Octane
First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
We have a solution of NaOH and H₂CO₃
First, NaOH will dissociate into Na⁺ and OH⁻ ions
The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃
The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule
If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃
