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Charra [1.4K]
3 years ago
13

Rocks, as they are compressed, begin forming mountains above the Earth's surface when two continental plates converge. The conti

nental crust increases in depth as the mountains grow above. The Himalayan Mountains formed at a convergent plate boundary in this manner. The rocks are smashed together causing them to __________ due to the intense heat and pressure from the colliding plates and eventually forming _____________ rock.
A) melt; igneous

B) layer; sedimentary

C) recrystallize; metamorphic

D) melt into the Earth's interior; metamorphic
Chemistry
1 answer:
Brrunno [24]3 years ago
7 0

The answer is; C

Due to the intense heat and pressure on the rocks, they undergo physical and chemical changes and become new types of rocks called metamorphic rocks. An example of this change is that small crystals may be squeezed into large crystals and the crystals may be rearranged.

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What is Mars colonization?
serious [3.7K]

Answer:The colonization of Mars has received interest from public space agencies and private corporations and has received extensive hypothetical treatment in science fiction writing, film, and art.

Explanation:

3 0
3 years ago
What combines to form compounds?
Citrus2011 [14]

Hello! Glad to help.

From what I have learned in my ICP class, molecules combine together to form componds.  A compound is any molecule that is made up of two or more different elemental atoms.

Hope this helps you! Anymore questions? Just ask.

6 0
4 years ago
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f
deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

           S = K_{b} \times ln \Omega

where,      S = entropy

             K_{b} = Boltzmann constant

             \Omega = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that \Omega = 1, and \Omega' = 0.833

Therefore, change in entropy will be calculated as follows.

     \Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega

                 = 1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)

                 = 1.38 \times 10^{-23} \times (-0.182)

                 = -0.251 \times 10^{-23}

or,             = -2.51 \times 10^{-24}

Thus, we can conclude that the entropy change for a particle in the given system is -2.51 \times 10^{-24} J/K particle.

8 0
3 years ago
A moose eats a lily pad. The food the moose eats is used to make ATP energy that can be used in cellular processes. The products
AURORKA [14]
Its is B (oxygen) and C (glucose)
6 0
3 years ago
Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th
Black_prince [1.1K]

Answer:

K_{goal}=1.793*10^{-33}

Explanation:

N2(g)+O2(g)⇌2NO(g), K_1 = 4.10*10^{-31}

N2(g)+2H2(g)⇌N2H4(g), K_2 = 7.40*10^{-26}

2H2O(g)⇌2H2(g)+O2(g), K_3 = 1.06*10^{-10}

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g)                     Eq (1)

Equilibrium constant for Eq (1) is K_1*K_3*K_3

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g)                       Eq (2)  

Equilibrium constant for Eq (2) is (K_1*K_3*K_3)^{1/2}

Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}

7 0
4 years ago
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