corinam9851 corinam9851
  • 02-04-2019
  • Chemistry
contestada

How many ml of 0.0500 m cacn2 are needed to make 25.0 ml of 0.0150 m solution? The molar mass of cacn2 is 80.11 g/mol.

Respuesta :

jamuuj jamuuj
  • 11-04-2019

Answer:

= 7.50 mL

Explanation:

m1v1=m2v2

In this case;

m1=0.05  

m2=0.015 and

v2=25.00mL  

Therefore;

(.0500M)(v1)=(25.00mL)(.0150M)

V1 = ((25.00mL)(0.0150M))/0.0500 M

    = 7.50 mL

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