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Sindrei [870]
3 years ago
10

If 500 mL of liquid mercury weighs 6.53 kg, what is the mass in pounds? A) 1.44 lb B) 2960 lb C ) 14.4 lb D) 7.19 lb E) 2.96 lb

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
8 0

Answer:

The correct option is: C ) 14.4 lb      

Explanation:

The standard International unit of mass is kilogram, kg. Other commonly used units of mass are grams (g), tonne (t), pounds (lb).

The mass of 500 mL liquid mercury is 6.53 kg.

Since, 1 kg = 2.20462 lb ≈ 2.205 lb

Therefore, the mass of 500 mL liquid mercury in pounds = 6.53 × 2.20462 = 14.39619 lb ≈ 14.4 lb.  

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Which of the following changes depending on the strength of the gravity field it is in?
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3 years ago
0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(
Studentka2010 [4]

Answer : The concentration of Fe^{3+} at equilibrium is 0 M.

Solution :  Given,

Concentration of Fe^{3+} = 0.0200 M

Concentration of C_2O_4^{2-} = 1.00 M

The given equilibrium reaction is,

                            Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)

Initially conc.       0.02         1.00                   0

At eqm.             (0.02-x)    (1.00-3x)                x

The expression of K_c will be,

K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}

1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}

By solving the term, we get:

x=0.02M

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Therefore, the concentration of Fe^{3+} at equilibrium is 0 M.

4 0
3 years ago
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