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3 years ago

Solve the following 2 questions below, involving elements and reactions. *

Chemistry

2 answers:

Korvikt [17]3 years ago

7 0

*Make proportions to solve. Find the rate from the question 1.

Question 1 is 40.

There are 400 grams of NaCI. There are 10 drops.

Divide 400 by 10.

400 / 10 = 40

40 is the answer.

Now all we do is repeat the same process for question 2.

Question 2 is 20.

There are 800 grams of barium peroxide, and there are 40 drops.

Divide 800 by 40.

800 / 40 = 20

Answer = 20.

padilas [110]3 years ago

5 0

Answer:

1. 40 grams of NaCI

2. 20 grams of Barium Peroxide

Explanation:

<h3><u>In this question, we have to divide to find the amount of each element in a drop. </u></h3>

For Question 1, there are 400 grams of NaCI. There are 10 drops.

So, we have to divide 400 by 10.

400 / 10 = 40

We get 40 as our result, so 40 is the answer.

<h3><u>Same for this question, we use the process we did before.</u></h3>

There are 800 grams of barium peroxide, and there are 40 drops.

So, we divide 800 by 40.

800 / 40 = 20

We have the answer as 20.

Our final answers are going to be listed below:

Question 1 - 40
Question 2 - 20

Hope this helps you!

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The Haber reaction for the manufacture of ammonia is: N2 + 3H2 → 2NH3 Without doing any experiments, which of the following can

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Answer : The correct statement is, $\text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

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

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

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.

The given rate of reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }N_2=-\frac{d[N_2]}{dt}$

$\text{Rate of disappearance of }H_2=-\frac{1}{3}\frac{d[H_2]}{dt}$

$\text{Rate of formation of }NH_3=+\frac{1}{2}\frac{d[NH_3]}{dt}$

From this we conclude that,

$\text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)$

Hence, the correct statement is, $\text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)$

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As the moisture-rich air approaches the mountain, it rises and cools, and water molecules in the air combine to form small drops

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<span>As the moisture-rich air approaches the mountain, it rises and cools, and water molecules in the air combine to form small drops. The process where water vapor becomes a liquid is called condensation. </span>

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Answer:

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For example,

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If we divide each side of the equation by 60 s, we get

$\dfrac{\text{1 min}}{\text{60 s}} = 1\\\\\text{Thus, both conversion factors, $\dfrac{\text{60 s}}{\text{1 min}}$ and $\dfrac{\text{1 min}}{\text{60 s}}$, equal 1.}$

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Answer:

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make P the subject of the equation

P = P'V'/V........... Equation 2

From the question,

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Substitute these values into equation 2

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