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3 years ago

!!PLS HELP!! How many molecules are in 25.4g Tricarbon tetrahydride?

Chemistry

2 answers:

katovenus [111]3 years ago

4 0

<h2>Explanation←</h2>

This means 17 g of ammonia gas is = 1 mole of ammonia gas molecule. Sence, there are 8.856 x 1023 8.856 x 10 23 molecules in 25.4g of ammonia...

Anwser: 0.1651251733326733

My Caculations 78% might be right

<h2>How Its Created--></h2>

Ammonia is produced naturally from decomposition of organic matter, including plants, animals and animal wastes. Also, BACTERIA? Didn't know

Well now human being you know!

<h2><em>Defenition Of Ammonia</em></h2>

Well The Definiton of Ammonia is colourless, pungent gas composed of nitrogen and hydrogen. ... It is the simplest stable compound of these elements and serves as a starting material for the production of many commercially important nitrogen compounds.

Elena-2011 [213]3 years ago

3 0

That would be 3907.09658

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Calcium forms ions with a charge of +2. Iodine forms ions with a charge of -1. Which of the following would represent an ionic c

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The function of the cilia and flagella are:

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The last one, to swim

Explanation:

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3 years ago

Which graph represents what happens to the solubility of baco3 as hno3 is added?

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If we have in mind the following procedure:
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3 years ago

What are the concentrations of A , A, B , B, and C C at equilibrium if, at the beginning of the reaction, their concentrations a

Elenna [48]

The question is incomplete, here is the complete question:

A reaction

$A+B\rightleftharpoons C$

has a standard free-energy change of -4.88 kJ/mol at 25°C

What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction their concentrations are 0.30 M, 0.40 M and 0 M respectively?

<u>Answer:</u> The equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

<u>Explanation:</u>

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_c$

where,

$\Delta G^o$ = Standard Gibbs free energy = -4.88 kJ/mol = -4880 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_c$ = equilibrium constant of the reaction

Putting values in above equation, we get:

$-4880J/mol=-(8.3145J/Kmol)\times 298K\times \ln K_c\\\\K_c=7.17$

We are given:

Initial concentration of A = 0.30 M

Initial concentration of B = 0.40 M

Initial concentration of C = 0 M

The chemical reaction follows:

$A+B\rightleftharpoons C$

<u>Initial:</u> 0.30 0.40 0

<u>At eqllm:</u> 0.30-x 0.40-x x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[C]}{[A][B]}$

We are given:

$K_c=7.17$

Putting values in above equation, we get:

$7.17=\frac{x}{(0.30-x)\times (0.40-x)}\\\\x=0.183,0.657$

Neglecting the value of x = 0.657, because change cannot be greater than the initial concentration

So, equilibrium concentration of A = $(0.30-x)=(0.30-0.183)=0.117M$

Equilibrium concentration of B = $(0.40-x)=(0.40-0.183)=0.217M$

Equilibrium concentration of C = $x=0.183M$

Hence, the equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

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