If you wish to prepare a 1.13 M solution of NaNo3, to what volume (in liters) would you have to dilute 25.4 mL of 3.21 M NaNO3?
1 answer:
Answer:
Answer is 72.15 or 0.07215
Explanation:
Given,
Initial molarity M1 = 3.21M
Initial volume V1 = 25.4mL
Final molarity of M2 = 1.13 M
Final volume V2 = ?
M1×V1=M2×V2
M1×V1/M2=V2
3.21M×25.4/1.13=V2
V2=72.15 mL. 1 mL=0.001 L
V2= 0.07215
You might be interested in
Answer:
use secondary data. the normal method to use
Answer:
nitrogen dioxide
Explanation:
this was on my chemistry test
2KClO3=2KCl+3O2
The equation
You thought this was a person but it's me dio
Because energy need to be Times 4. Not sure if I helped you
