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cluponka [151]
3 years ago
6

A copper block (20°C) is placed in contact with a lead block (30°C), which is already in contact with an iron block (70°C). What

will happen in this situation? Heat will flow from the copper to the lead and from the iron to the lead until the temperatures are equal. Heat will flow from the lead to the copper and to the iron until the temperatures are equal. Heat will flow from the iron to the lead and from the lead to the copper until the temperatures are equal. Heat will flow from the copper to the lead to the iron until the temperatures are equal.
Chemistry
1 answer:
dlinn [17]3 years ago
8 0

Answer:

The heat flows from iron block to lead block then it flows from lead block to iron block until the temperature of three blocks are equal.

Explanation:

Given data

Temperature of the copper block = 20 °c

Temperature of the lead block = 30 °c

Temperature of the iron block = 70 °c

We know that according to law of thermodynamics heat always flows from higher temperature body to lower temperature body Until the temperature of all bodies becomes equal.

Therefore in the given problem the heat flows from iron block to lead block then it flows from lead block to iron block until the temperature of three blocks are equal.

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

Solution :  Given,

Concentration of Fe^{3+} = 0.0200 M

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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

Concentration of Fe^{3+} at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

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

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The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s
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Answer : The equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

Explanation :

The dissociation of acid reaction is:

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Initial conc.        c                                 0                0

At eqm.             c-x                                 x                x

Given:

c = 7.0\times 10^{-2}M

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The expression of dissociation constant of acid is:

K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}

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