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4 years ago

What is 2 million divided by 8

Chemistry

2 answers:

geniusboy [140]4 years ago

7 0

Answer:

use a calculator please

Flura [38]4 years ago

6 0

Answer:

250,000

Explanation:

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Problem 1: For which of the following activities might you want to hire a chemist?

Tatiana [17]

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3 years ago

What is acid precipitation?

zimovet [89]

Explanation:

precipitation like snow rain with increased acid is know as acid precipitation

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3 years ago

Which type of reaction is the Haber process: N2(g) + 3 H2(g) → 2 NH3(g) + heat? *

Alik [6]

Answer:

exothermic, with a decrease in entropy

Explanation:

Whenever you produce heat as a product in a reaction, the reaction is exothermic. To determine entropy, we know we have 4 moles of gas on reactant (1 from N2 and 3 from H2) and in produce side we only have two moles (2 from NH3) thus since we are decreasing the number of gas molecules, there is going to be less disorder, hence decrease in entropy.

8 0

3 years ago

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 :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

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.

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4 years ago

On a hot summer day, air rises most quickly when it is above a

GalinKa [24]

I’d say probably a forest

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3 years ago