The slight positive charges on the hydrogen atoms in water molecules attract the slight negative charges on the oxygen atoms of the other water molecules
Answer:
The answer to your question is Molarity = 0.0708
Explanation:
Data
NaOH 0.05 M Volume 1 = 3.87 ml Volume 2 = 25.11 ml
HCl 15 ml
Process
1.- Find the volume used of NaOH
25.11 - 3.87 = 21.24 ml = 0.02124 l
2.- Write the balanced equation of the reaction
NaOH + HCl ⇒ NaCl + H₂O
3.- Calculate the moles of NaOH in the solution
Molarity = 
moles = Molarity x volume
moles = 0.05 x 0.02124
moles = 0.001062
4.- From the reaction we know that NaOH and HCl react in a proportion 1:1.
1 mol of NaOH ------------- 1 mol of HCl
0.001062 moles of NaOH ------------ x
x = (0.001062 x 1) / 1
x = 0.001062 moles of HCl
5.- Find the molarity of HCl
Molarity = 
Molarity = 0.0708
The charge of Br changed from –1 to 0, therefore it is the
element which is oxidized. Since it is oxidized then Br is also the reducing
agent.
The charge of Mn changed from +4 to +2 therefore it is the
element which is reduced. Since Mn is reduced, then MnO2 is the oxidizing
agent.
The half –reactions are:
Br: 2Br --> Br2 + 2e-
Mn: MnO2 --> Mn2+
First balance oxygen by adding H2O:
MnO2 --> Mn2+ + 2H2O
Then balance hydrogen by adding H+ ions:
4H+ + MnO2 --> Mn2 + 2H2O
Then the appropriate electrons:
4e- + 4H+ + MnO2 --> Mn2 + 2H2O
Multiply the half-reaction of Br by 2 because the half-reaction
of Mn has 4 electrons.
4Br --> 2Br2 + 4e-
Combine the two half reactions and cancel common factors:
4Br- + 4H+ + MnO2 --> 2Br2 + Mn2 + 2H2O
Answer:
A. 1.63g/dm^3 (3 s.f.)
B. 0.833g/dm^3 (3 s.f.)
C. 1.92g/dm^3 (3 s.f.)
Explanation:
Please see attached picture for full solution.