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!
Answer:
Sorry; this looks like some specific point from class or your book. I can think of each one and create a synthesis, but that isn’t what this is.
Explanation:
Answer:
i. CH2 = C(CH3)CH=C(Cl) - CH3
<u> 4-methyl pent-2,4-diene-2-chloro</u>
Explanation:
ii.CH3 CH2 C(CH3) Cl CH(CH3) C(Br)2 CH3
2-Bromo -4 chloro-2,3,4 tri methyl hexane
The safest method for diluting
concentrated sulfuric acid with water is to add acid to water. This way, when
spill occurs, the acid is already diluted and less harmful than adding water to
acid.
