The concentrations of Fe and K in a sample of riverwater are 0.0400 mg/kg and 1.30 mg/kg, respectively. Express the concentration in molality.

Answer :

The concentration in molality of Fe is,

The concentration in molality of K is,

Explanation:

First we have to calculate concentration in molality of Fe.

Molar mass of Fe = 56 g/mol

Concentration of Fe = 0.0400 mg/kg =

Conversion used : 1 g = 1000 mg

Thus, the concentration in molality of Fe is,

Now we have to calculate concentration in molality of K.

Molar mass of K = 39 g/mol

Concentration of K = 1.30 mg/kg =

Conversion used : 1 g = 1000 mg

Thus, the concentration in molality of K is,

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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.