Vanadium has two naturally occurring isotopes, 50V with an atomic mass of 49.9472 amu and 51V with an atomic mass of 50.9440. The atomic weight of vanadium is 50.9415. The percent abundances of the vanadium isotopes are % 50V and % 51V.

The percent abundances of the vanadium isotopes are 0.25% 50V and 99.75% 51V.

Explanation:

Let 50V be isotope A

Let 51V be isotope B

Mass number of isotope A (50V) = 49.9472

Mass number of isotope B (51V) = 50.9440

Abundance of isotope A (50V) = A%

Abundance of isotope B (51V) = B% = 100 — A

The atomic weight of vanadium

= 50.9415

Atomic weight = [(Mass of A x Abundance of A)/100] + [(Mass of B x Abundance of B) /100]

50.9415 = [(49.9472xA%)/100] + [(50.9440x(100 — A))/100]

50.9415 = [49.9472A%/100] + [(5094.40 — 50.9440A%)/100]

Multiply through by 100

5094.15 = 49.9472A% + 5094.40 — 50.9440A%

Collect like terms

5094.15 — 5094.40 = 49.9472A% — 50.9440A%

—0.25 = — 0.9968A%

Divide both side by — 0.9968

A% = —0.25/ — 0.9968

A% = 0.25%

B% = 100 — 0.25 = 99.75%

The percent abundances of the vanadium isotopes are 0.25% 50V and 99.75% 51V.

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"