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klemol [59]
2 years ago
13

PLEASE HELP ME! THANK YOU IF YOU DO!!! ^^

Chemistry
1 answer:
frozen [14]2 years ago
4 0

Answer:

oceanic formation is the right answer.

Explanation:

this os becoz they slide a past each other and do not rub against each other

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Which of the following would not be an isotope of carbon-12? A. Carbon-14 B. Carbon-13 C. Boron-12 D. Carbon-11
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C because it isn't carbon
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You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th
ASHA 777 [7]

Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially,  only 4 moles of oxygen gas were present in the flask.

p_{O_2}=Tp_1\times X_{O_2}  (X_{O_2}=\frac{4}{4}) ( according to Dalton's law of partial pressure)

p_{O_2}=Tp_1\times 1=Tp_1....(1)

Tp_1= Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

Tp_2= Total pressure of the mixture.

from (1)

Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

On rearranging, we get:

Tp_2=2\times Tp_1

The new total pressure will be twice of initial total pressure.

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erastovalidia [21]

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The correct answer would be C) Support

4 0
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Which of these is NOT a falsifiable hypothesis?
pashok25 [27]
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2 years ago
Read 2 more answers
CdF2(s)⇄Cd2+(aq)+2F−(aq)A saturated aqueous solution of CdF2 is prepared. The equilibrium in the solution is represented above.
kupik [55]

Answer:

The correct answer is option a.

Explanation:

CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)

Equilibrium concentration cadmium ions = [Cd^{2+}]=0.0585 M

Equilibrium concentration fluoride ions = [F^{-}]=0.117 M

Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.

The molar solubility of the solid cadmium fluoride = 0.0585 M

CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)..[1]

NaF(s)\rightleftharpoons Na^{+}(aq)+F^-(aq)

Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.

Hence, decrease in solubility will be observed.

8 0
2 years ago
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