A35.00 ml sample of mg(no3)2 was diluted with water to 62.50 ml. a 20.00 ml sample of the dilute solution contained 0.300 m no3−(aq). what was the concentration of mg(no3)2 in the original, undiluted solution? hint:

0,268M

Explanation:

A dilution is a process in which the concentration of a solution is reduced by adding more solvent.

In the problem, the sample of Mg(NO₃)₂ was diluted from 35,00mL to 62,50mL -dilution factor: 62,50mL/35,00mL- . The NO₃⁻ concentration of this diluted solution was 0,300M, as Mg(NO₃)₂ has two NO₃⁻, the concentration of Mg(NO₃)₂ in the diluted solution is:

0,300M NO₃⁻× = 0,150M Mg(NO₃)₂

Using dilution factor, the original concentration of Mg(NO₃)₂ is:

0,150M Mg(NO₃)₂× = 0,268M

I hope it helps!

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles

am i too late?

explanation: