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atroni [7]
3 years ago
10

Water circulates through Earth's water cycle by changing state state its acidity its acidity its turbidity

Chemistry
1 answer:
Lana71 [14]3 years ago
3 0

Complete question is;

Water circulates through Earth's water cycle by changing its _____.

A. state

B. acidity

C. turbidity

D. composition

Answer:

A: State

Explanation:

We can talk about this by observations about the states of water in the Earth's water cycle and even nature.

For example, water changing from liquid to steam(gas) in evaporation; under certain temperatures/conditions snow undergoes sublimation directly from solid to gas and also undergoes melting from solid to liquid; water vapour in the atmosphere undergoes condensation and cooling to form rain.

Thus, we can say that water circulates through the Earth's water by changing it's state.

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75% (TT,Tt,Tt,tt) hope that helps :)
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What forms when all of the continents on earths surface merge into one massive landform?
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Hope this helped.
6 0
4 years ago
What is the mass of 6.12 moles of arsenic (As)?
Leokris [45]

Mole = Mass / Molar mass

6.12 moles = Mass / 74.92 g/mol

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7 0
3 years ago
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In the balanced equation , CS2 + 3O2 = CO2 + 2SO2 , how many mol of O2 would react with 34.5 mol of CO2?
Arada [10]

<u>Given:</u>

Moles of CS2 (it cannot be CO2 as mentioned in the question, since O2 reacts with CS2 and not CO2) = 34.5 mol

<u>To determine:</u>

Moles of O2 undergoing the reaction

<u>Explanation:</u>

The reaction is-

CS2 + 3O2 → CO2 + 2SO2

Based on the stoichiometry: 3 moles of O2 reacts with 1 mole of CS2

therefore the moles of O2 that would combine with 34.5 moles of CS2 are

= 3 moles O2 * 34.5 moles CS2/1 mole CS2 = 103.5 moles

Ans: Around 104 moles of O2 would react with 34.5 moles of CS2





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3 years ago
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Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

Here A and A cancel each other

B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

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1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

7 0
3 years ago
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