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disa [49]
3 years ago
6

Vanadium has an atomic mass of 50.9415 amu. It has two common isotopes. One isotope has a mass of 50.9440 amu and a relative abu

ndance of 99.75 %.
What is the abundance of the other isotope?
What is the mass of the other isotope?
Chemistry
1 answer:
grandymaker [24]3 years ago
6 0

Answer:

Average atomic mass of the  vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

Explanation:

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See explaination

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When 1.95 g of co(no3)2 is dissolved in 0.350 l of 0.220 m koh, what are [ co2+], [ co(oh)42−], and [ oh−] if kf of co(oh)42− =
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V KOH = 0.350 L
[KOH] = 0.220 M
Kf = 5.0 x 10⁹
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so [Co(NO₃)₂] = 1.95 / (0.350 * 182.943) = 0.03045 M
[Co²⁺] = 0.03045 M
[OH⁻] = 0.22 M
chemical reaction:
             Co²⁺(aq) + 4 OH⁻    ⇄      Co(OH)₄²⁻
I (M)      0.03045      0.22                   0
C (M)   - 0.03045   - 4 (0.03045)      0.03045
E (M)       - x         0.22 - 4(0.03045)   0.03045
                              = 0.0982
Kf  = [Co(OH)₄²⁻] / [Co⁺²][OH⁻]⁴
5.0 x 10⁹ = (0.03045) / x (0.0982)⁴
x = 6.5489 x 10⁻⁸
at equilibrium:
[Co²⁺] = 6.54 x 10⁻⁸
[OH⁻] = 0.0982 M
[Co(OH)₄²⁻] = 0.03045 M



4 0
3 years ago
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