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gogolik [260]
2 years ago
14

Hi guys can someone help please I didn’t this three time already but I still got it wrong can someone please give me the answers

or examples sorry guys I have two more I just need help

Chemistry
2 answers:
kirill115 [55]2 years ago
7 0

Answer:

━━━━━━━━━━━━━━━━━━━━━━━━

Q.17 7.23*10²⁴ particles of Cl2 = 12 moles of Cl2

━━━━━━━━━━━━━━━━━━━━━━━━

Q.18 2.89*10²⁴ molecules of water = 4.8 moles of water

━━━━━━━━━━━━━━━━━━━━━━━━

Q.19 7.95*10²³ molecules of C2F2H4 = 1.32 moles of C2F2H4

━━━━━━━━━━━━━━━━━━━━━━━━

Q.20 1.45*10²⁴ molecules of AgCl = 2.4 moles of AgCl

━━━━━━━━━━━━━━━━━━━━━━━━

<em><u>Thanks for joining brainly community!</u></em>

koban [17]2 years ago
4 0
1.2 cl2 mol

4.8 water mol

1.3 mol

2.4 Agcl mol
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Consider a simple reaction in which a reactant AA forms products: A→productsA→products What is the rate law if the reaction is z
levacccp [35]

Answer :

The rate law expression for zero order reaction will be:

Rate=k[A]^0

The rate law expression for first order reaction will be:

Rate=k[A]^1

The rate law expression for second order reaction will be:

Rate=k[A]^2

Zero order reaction : There is no affect on the rate law.

First order reaction : The rate law becomes doubled.

Second order reaction : The rate law becomes quadrupled.

Explanation :

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The given reaction is:

A\rightarrow Products

The rate law expression for zero order reaction will be:

Rate=k[A]^0

The rate law expression for first order reaction will be:

Rate=k[A]^1

The rate law expression for second order reaction will be:

Rate=k[A]^2

Now we have to determine that if doubling of the concentration of A then the rate of reaction will be:

As we know that the zero order reaction does not depend on the concentration of reactant. So, there is no affect on the rate law.

As we know that the first order reaction depend on the concentration of reactant. So, the rate law becomes doubled.

As we know that the second order reaction depend on the concentration of reactant. So, the rate law becomes quadrupled.

5 0
4 years ago
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