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

Answer Link

Otras preguntas

Which tool might be used to limit imports from Cuba but maintain exports to Cuba? a. embargo b. NAFTA c. sanctions d. restricted trade routes
Help? I don't know how to do this.
Change into indirect speech and please explain the rules also. Thank you so much.
PLEASE ANSWER FAST!!!!!! Question 7 Multiple Choice Worth 4 points) (01.03 MC) Object A has a density of 10 g/cm and a mass of 9 g. Object B has a density of 10
[tex]A=\frac{1}{2}\pi w^2+2lw[/tex]Solve for L
Find the perimeter of the shape. Assume all intersections are right angles. A) 8.5 B) 10 + 3 √2 C) 13 D) 13 + 2 √3 E) 15
Which of these groups was most likely to oppose the colonists independence from Britain
The perimeter of a triangle is 137 cm. If three sides of the triangle are x + 7cm, 2x - 5 and 3x - 3 cm, what is the length of each side ?
What are effectiveness and ineffectiveness of road user??
Which one of the base sequences are accurately paired? G with C G with A T with C C with T