If you prepare 525 mL of a .650M MgCl2 solution, how many grams of solute would you need?

Explanation:

525ml of 0.6550M MgCl₂ => 0.525L x 0.650M = 0.3413mole MgCl₂

0.3413mole MgCl₂ = 0.3413mole x 94 g/mole = 32.1g MgCl₂

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.

a:

by the combination of an unlimited number of elements