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belka [17]
3 years ago
6

Two identical spoons are electroplated with Ag or Cd through the use of the electrolytic cells. A current of 5.00A was supplied

to each cell for 600. seconds, and the masses of the spoons before and after the electroplating were recorded. Write down the mathematical equations can best be used to account for the much larger increase in mass of the spoon electroplated with Ag compared with the spoon electroplated with Cd.
Chemistry
1 answer:
OLga [1]3 years ago
4 0

Answer:

Explanation:

One farad of charge is capable of depositing one gram equivalent of a metal

One gram equivalent of Ag = 108 grams

One gram equivalent of Cd = 112 / 2  grams

= 56 grams . [ for cadmium equivalent mass = atomic mass / s ]

electric charge flowing = current x time = 5 x 600 = 3000 coulomb

one farad = 96500 coulomb

96500 coulomb  deposits 108 gram of Ag

3000 coulomb deposits 108 x 3000 / 96500 gram

= 3.35 grams of Ag

Similarly ,

96500 coulomb  deposits 56 gram of Cd

3000 coulomb deposits 56 x 3000 / 96500 gram

= 1.74  grams of Cd

So there will be much larger increase in the spoon of Ag due to larger deposit of Ag by charge .

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The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.
Brilliant_brown [7]

Answer :  The rate of reaction is,

Rate=4.77\times 10^{-19}M/s

The appearance of NO_3 is, 4.77\times 10^{-19}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)

The rate law expression will be:

Rate=k[NO_2][O_3]

Given:

Rate constant = k=1.69\times 10^{-4}M^{-1}s^{-1}

[NO_2] = 1.77\times 10^{-8}M

[O_3] = 1.59\times 10^{-7}M

Rate=k[NO_2][O_3]

Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})

Rate=4.77\times 10^{-19}M/s

The expression for rate of appearance of NO_3 :

\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}

As, \text{Rate of reaction}=4.77\times 10^{-19}M/s

So, \text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s

Thus, the appearance of NO_3 is, 4.77\times 10^{-19}M/s

7 0
3 years ago
A 9.79 mol sample of freon gas was placed in a balloon. Adding 3.50 mol of freon gas to the balloon increased its volume to 21.8
aivan3 [116]

Answer:

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Explanation:

Initial moles of freon in ballon = n_1=9.79 mol

Initial volume of freon gas in ballon = V_1=?

Moles of freon gas added in the balloon = n = 3.50 mole

Final moles of freon in ballon = n_2=n_1+n=9.79 mol+3.50 mol=13.29 mol

Final volume of freon gas in ballon = V_2=21.8 L

Using Avogadro's law:

\frac{V_1}{n_1}=\frac{V_2}{n_2} ( at constant pressure and temperature)

V_1=\frac{V_2\times n_1}{n_2}=\frac{21.8 L\times 9.79 mol}{13.29 mol}=16.06L

16.06 L was the initial volume of the balloon.

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Keith_Richards [23]

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