Answer: a) 
b) 1 mole of 
is produced.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The skeletal equation is:
The balanced equation will be:
Thus the coefficients are 2, 3 , 10 , 4 , 3 , 2 and 5.
b) Oxidation: 
Reduction: 
Net reaction: 
When 1 mole of 
is produced, 1 mole of is produced.
During a phase change the temperature does not change since all of the heat is being absorbed in order to break the intermolecular forces. Due to that, the formula will not need to have T in it and is actually q=nΔH(v).
n=the number of moles (in this case 2.778mol of water since you divide 50g by 18g/mol).
ΔH(v)=the molar heat of vaporization (in this case 40.7kJ/mol).
q=the heat that must be absorbed
q=2.778mol×40.7kJ/mol
q=113.1kJ
Therefore the water needs to absorb 1.13×10²kJ.
I hope this helps. Let me know if anything is unclear.
Answer:
HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)
Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:
moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl
moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH
You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC
Answer:
-24.76 kJ/g; -601.8 kJ/mol
Explanation:
There are two heat flows in this experiment.
Heat from reaction + heat absorbed by calorimeter = 0
q1 + q2 = 0
mΔH + CΔT = 0
Data:
m = 0.1375 g
C = 3024 J/°C
ΔT = 1.126 °C
Calculations:
0.1375ΔH + 3024 × 1.126 = 0
0.1375ΔH + 3405 = 0
0.1375ΔH = -3405
ΔH = -24 760 J/g = -24.76 kJ/g
ΔH = -24.76 kJ/g ×24.30 g/mol = -601.8 kJ/mol
First determine the formal oxidation numbers:
N changes from +2 to +5 going from NO to (NO3)- O remains -2 the whole time Cr changes from +6 to +3
Now write the half reactions, balance the oxygens with the required number of waters and then balance the hydrogens with the required number of protons:
Oxidation half reaction:
NO(aq) + 2 H2O(l) ---> (NO3)-(aq) + 4 H+(aq) + 3 e-
Reduction half reaction:
(Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l)
Now balance the number of electrons on both sides and add them together:
2 NO(aq) + 4 H2O(l) ---> 2 (NO3)-(aq) + 8 H+(aq) + 6 e- (Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l) --------------------------------------... 2 NO(aq) + (Cr2O7)2-(aq) + 6 H+(aq) ---> 2 (NO3)-(aq) + 2 Cr3+(aq) + 3 H2O(l)
Notice that the charge is the same in both sides, which is an indication that the redox equation has been balanced correctly:
-2 + 6 = -2 + 2(+3) +4 = +4
