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3 years ago

The cracking of rock due to heat is an example of

Chemistry

2 answers:

lora16 [44]3 years ago

6 0

c is wrong it’s physical weathering

Advocard [28]3 years ago

3 0

Answer: chemical weathering

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Plz help with this it is the digestive system

Strike441 [17]

Answer:

see explanation for answer

Explanation:

salivary gland: 1

stomach: 2

small intestine: 6

liver: 4

gallbladder: 5

large intestine: 3

The answers correspond with the numbers on the text boxes, so you would drag number 1 to the salivary gland and so on.

5 0

3 years ago

Read 2 more answers

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

What is the mass of solute in a 500 mL solution of 0.200 M

Fofino [41]

16.4 grams is the mass of solute in a 500 mL solution of 0.200 M .

sodium phosphate

Explanation:

Given data about sodium phosphate

atomic mass of Na3PO4 = 164 grams/mole

volume of the solution = 500 ml or 0.5 litres

molarity of sodium phosphate solution = 0.200 M

The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:

The formula is

molarity = $\frac{number of moles of solute}{volume in litres}$

putting the values in the equation, we get

molarity x volume = number of moles

0.200 X 0.5= number of moles

number of moles = 0.1 moles

Atomic mass x number of moles = mass

putting the values in the above equation

164 x 0.1 = 16.4 grams

16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.

8 0

3 years ago

Como se escribe K2O ennomemclatura estequimetrica y stock

Paladinen [302]

i speak english . gracias!

4 0

3 years ago

How is an oscilloscope used to tune a musical instrument?

sp2606 [1]

It uses the voltages and sound freq. in the air to measure the wave lengthths

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3 years ago