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r-ruslan [8.4K]
3 years ago
6

Find the sum of - 1 /5 and reciprocal of 6 /11 ​

Chemistry
1 answer:
Bess [88]3 years ago
7 0

<em>Answer:</em>

<h2><em>4</em><em>9</em><em>/</em><em>3</em><em>0</em></h2>

<em>Solution</em><em>,</em>

<em>Reciprocal</em><em> </em><em>of </em><em>6</em><em>/</em><em>1</em><em>1</em><em>=</em><em>1</em><em>1</em><em>/</em><em>6</em>

<em>Now,</em>

<em>-</em><em>1</em><em>/</em><em>5</em><em>+</em><em>1</em><em>1</em><em>/</em><em>6</em>

<em>=</em><em>-</em><em>1</em><em>*</em><em>6</em><em>+</em><em>1</em><em>1</em><em>*</em><em>5</em><em>/</em><em>3</em><em>0</em>

<em>=</em><em>-</em><em>6</em><em>+</em><em>5</em><em>5</em><em>/</em><em>3</em><em>0</em>

<em>=</em><em>4</em><em>9</em><em>/</em><em>3</em><em>0</em>

<em>Hope </em><em>it</em><em> helps</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

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The expression for rate of reaction will be :

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

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\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,

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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.

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Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}

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\\ \large\sf\longmapsto KNO_3

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