본문 바로가기
화학

질산 용액의 노르말농도. 27.5 mL 3.5×10^-2 N Ca(OH)2 + 10.0 mL HNO3

by 영원파란 2020. 6. 23.

AdSense@ywpop

728x170

질산 용액의 노르말농도. 27.5 mL 3.5×10^-2 N Ca(OH)2 + 10.0 mL HNO3

 

 

If 27.5 mL of 3.5×10^-2 N Ca(OH)2 solution is needed to neutralize 10.0 mL of nitric acid solution of unknown concentration, what is the normality of the nitric acid?

 

---------------------------------------------------

 

NV = N’V’

( 참고 https://ywpop.tistory.com/4689 )

 

 

 

(? N) (10.0 mL) = (3.5×10^(-2) N) (27.5 mL)

? = (3.5×10^(-2)) (27.5) / (10.0) = 0.09625 N

 

 

 

: 9.63×10^-2 N

 

 

 

 

[키워드] HNO3 10.0 mL 3.5×10^-2 N Ca(OH)2 27.5 mL

 

 

반응형
그리드형(광고전용)

댓글