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nexus9112 [7]
3 years ago
10

3(x - 2) = 5(x + 4)​

Chemistry
2 answers:
MatroZZZ [7]3 years ago
8 0

Answer:

The answer is x = -13 .

Explanation:

Solve the equation:

3(x -2) = 5(x + 4)

Use <u>Distributive Property</u>:

3(x -2) = 5(x + 4)

3x - 6 = 5x + 20

-Subtract 5x from 3x :

3x - 6 -5x = 5x -5x + 20

-2x - 6 = 20

-Add both sides by 6 :

-2x - 6 + 6 = 20+6

-2x = 26

-Divide both sides by 2 :

\frac{-2x}{-2} = \frac{26}{-2}

x = -13

So, now you have found the answer.

Tcecarenko [31]3 years ago
5 0

Answer:

Uh first of all this is algebra but I'll answer this

First distribute the three and 5 (Multiply them by both terms inside parenthesis.

3x-6=5x+20

Then add like terms

8x=14

Divide 8 by 8 and 8 by 14

x = 14/8

Explanation:

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<u>Explanation:</u>

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For the given chemical reaction:

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The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol

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Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = \frac{121.77}{1}\times 1.90=231.36 J/K

Hence, the value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

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