From the chemical equation given:
H2SO4+2KOH--->K2SO4+2H2O
the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.
No. of moles of KOH = 2* no. of moles of H2SO4
=2*0.1*0.033
The concentration of KOH = no. of moles / volume
=2*0.1*0.033/0.05
=0.132M
a) 
→ 
b) 
→ 
<h3>
What are half-reactions?</h3>
The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.
a)
→
= 
= -0.83 - (-2.71) =1.88V
b)
→
= 
=-0. - (0.8) =-0.8V
Learn more about the half-reactions here:
https://brainly.in/question/18053421
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To determine this, lets use the atomic mass for each element to determine the corresponding number of moles for a given mass of 1 g.
Mo: 95.94 g/mol
1 g/95.94 g/mol = 0.01042 moles Mo
Se: 78.96 g/mol
1 g/78.96 g/mol = 0.01266 moles Se
Na: 22.99 g/mol
1 g/22.99 g/mol = 0.0435 moles Na
Br: 79.9 g/mol
1 g/79.9 g/mol = 0.0125 moles Br
<em>Thus, the answer is Na.</em>
