How many liters of 1.5 M potassium permanganate could be made if 152 g of the solute are available?

0.64 L

Explanation:

Recall that

n= CV where n=m/M

Hence:

m/M= CV

m= given mass of solute =152g

M= molar mass of solute

C= concentration of solute in molL-1 = 1.5M

V= volume of solute =

Molar mass of potassium permanganate= 158.034 g/mol

Thus;

152 g/158.034 gmol-1= 1.5M × V

V= 0.96/1.5

V= 0.64 L

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.