To prepare 250mL of calcium chloride solution with a molar concentration of 1.20mol/L, what mass of calcium chloride would be required?

33.30 grams of CaCl2 will be required

Explanation:

Given,

Volume of solution, V= 250 ml

Molarity of solution, M= 1.20 mol/L

Molecular mass of CaCL2, S= 40+(35.5 X 2)= 111

We know,

Required mass, W= SVM/1000

Now,

W = (111 X 250 X 1.20)/1000

    = 33300/1000

    = 33.30

Therefore, 33.30 grams of Calcium Chloride will be required.

your picture is not part of the scientific method.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.