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3 years ago

What is the difference between creep and a landslide

1 answer:

5 0

Mass movement is a

fast process

in the formation of landslides.

Mass movement is a

slow process

in the formation of creep.

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nadya68 [22]

Answer:

1 Hour...

Explanation:

If It’s 960 Km/h, then it will take 1 hour to fly there.

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2 years ago

A chemical reaction that takes in heat is called: exothermic endothermic a phase change

AlladinOne [14]

It would be Endothermic.

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2 years ago

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zepelin [54]

I would say G sorry if it’s not right

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3 years ago

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If you have a 4.6 L of gas in a piston at a pressure of 1.2 atm and compress 2p the gas until its volume is 2.6 L, what will the

Nesterboy [21]

Answer:

2.12atm

Explanation:

Boyle's Law: P1V1 = P2V2

Manipulate to solve for unknown: P2 = P1V1/V2

Substitute values: P2=(1.2atm)(4.6L)/2.6L

P2 = 2.1230769atm

Round to 3 sig figs to get 2.12atm

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2 years ago

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

3 years ago