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balandron [24]
3 years ago
8

How many moles are contained in 3.131 × 1024 particles?

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
5 0
One mole contains 6.02×10²³
If x mole contains 3.131×10²⁴

x = 3.131×10²⁴
-------------
6.02×10²³

x = 0.52×10²⁴¯²³

x = 0.52×10¹

x = 5.2moles....

Hope this helped...?
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This question uses mole ratios. firstly, we look at how many moles of H2 are produced for every mole of HCl: from 1 Hcl, we get half a mole of H2. therefore, however many moles of HCl are present in the 0.075 L of 1.0M HCl, half that number of H2 moles will be produced. 

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3 0
3 years ago
Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.84×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.14×10−4, what is the
Nimfa-mama [501]

Answer: The value of equilibrium constant for reaction is, 1.42\times 10^{-2}

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(2) AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq) ;  K_4=1.14\times 10^{-4}

Now we have to calculate the equilibrium constant for chemical equation as:

PbCl_2(aq)+2Ag^{+}(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq) ;  K=?

We are reversing reaction 2 and multiplying reaction 2 by 2 and then adding both reaction, we get the final reaction.

The equilibrium constant for the reverse reaction will be the reciprocal of that reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of that reaction will be the square of the equilibrium constant.

If we are adding equations then the equilibrium constants will be multiplied.

The value of equilibrium constant for reaction is:

K=(\frac{1}{K_4})^2\times K_3

Now put all the given values in this expression, we get:

K=(\frac{1}{1.14\times 10^{-4}})^2\times (1.84\times 10^{-10})

K=1.42\times 10^{-2}

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Answer:

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3 0
3 years ago
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