The naturally occurring isotopes of vanadium are vanadium-50 and vanadium-51. vanadium-50 has an abundance of 0.250% and a mass of 49.947 amu. vanadium-51 has an abundance of 99.750% and a mass of 50.944 amu. calculate the average atomic mass of vanadium.
a) 50.942 amu
b) 62.558 amu
c) 63.303 amu
d) 5094.151 amu

The answer to your question is letter A

Explanation:

                            Abundance              Atomic mass

       V-50             0.25%                         49.947

       V-51            99.750%                       50.944

Formula

Average atomic mass = [V-50-Abundance x V-50-Atomic mass] +

                                       [ V-51-Abundance x V-51-Atomic mass]

Substitution

Average atomic mass = [ 0.0025 x 49.947] + [0.9975 x 50.944]

                                    = 0.125 + 50.817

Result

                                    = 50.942 amu

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

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.