Hey there!:
Concentration of NaOH = 0.200 M
Concentration of HNO₃= 0.200 M
Total volume = 50.0 mL + 60.0 mL = 110 mL=> 0.11 L
The neutralization reaction between NaOH and HNO3 :
OH⁻ + H⁺ ----------> H₂O
So :
n ( H⁺ ) = 60 mL * 0.200 M / 1000 mL => 0.012 moles of H⁺
n ( OH⁻ ) = 50 mL 0.200 M / 1000 mL => 0.01 moles of OH⁻
Hence OH⁻ is limiting reagent .
Remaining moles of H⁺ = 0.012 - 0.01 => 0.002 moles
Concentration of H⁺ = 0.002 M / 0.11 L
Concentration of H⁺ = 0.01818 moles/L
Therefore:
pH = - log [ H⁺ ]
pH = - log [ 0.01818 ]
pH = 1.74
Hope that helps!
Add the reltive atomic mass of Ca and O which are 40 and 16 respectively after addition we get molecular mass of CaO which is 40+16=56 now divide atomic mass of O by molecular mass of CaO and multiply it with 100 as show n 16/56 *100=28.57%
Answer - B Atoms in the original substances are arranged in a different way to make new substances
It’s will decrease I have to make it 20 characters but it’s decrease
Answer:
The volume of NaOH required is - 0.01 L
Explanation:
At equivalence point
,
Moles of 
= Moles of NaOH
Considering
:-
Given that:
So,
<u>The volume of NaOH required is - 0.01 L</u>
