Stock solutions of nitric acid have concentration of 16.0 M. What volume of stock solution is needed to prepare 2.00 L of 2.00 M solution of nitric acid by dilution with water?
500 mL
125 mL
250 mL
32.0 mL
62.5 mL

250 mL

Explanation:

These are the variable one has to have in mind:

= 16 M

= 2 M

= 2 L = 2000 mL

Note: C = concentration; V = volume; subscript 1 = stock solution; subscript = 2 = dilution.

What we have to solve is for

Use the following equation:

x = x ⇒ = ⇒   = 250 mL

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.

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