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3 years ago

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Within the rock cycle, a metamorphic rock could become an igneous rock by going through which steps and in what order?

2 answers:

Yuki888 [10]3 years ago

5 0

Hello,

Question- Within the rock cycle, a metamorphic rock could become an igneous rock by going through which steps and in what order?

A. melting, cooling, crystallization

B. crystallization, melting, cooling

C. burial, compaction, cementation

D. weathering, transportation, burial

Answer- Your answer is A. Melting, cooling, crystallization.

Why- Any rock can become any rock igneous metamorphic or even condemnatory just by heating up but when it melts it becomes lava and the rock cycle starts all over again. See all it has to do to become a new phase of rock is two steps melt and cool and boom its a new phase of rock.

Important- If my answer was enough to help uou please mark me as Brainliest, Leave a thanks, and Rate my answer 5 stars thank you and have the best day ever!

swat323 years ago

3 0

Answer:

a is the right anwser it just makes sence

Explanation:

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If the density of an ideal gas at stp if found to be 0.0902 g/l, what is its molar mass?

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In the following overall chemical reaction, aluminum displaces chromium from chromium(III) oxide and forms aluminum oxide.

mafiozo [28]

Answer:

Change in enthalpy for the reaction is -536 kJ

Explanation:

Overall chemical reaction can be represented a summation of two given elementary steps with slight modification.
Take reaction (1a) and divide stoichiometric coefficients by 2
Take reverse reaction (2a) and divide stoichiometric coefficient by 2
Then add these two modified elementary steps to get overall chemical reaction
$\Delta H$ is an additive property. hence value of $\Delta H$ will be changed in accordance with modification

$2Al+\frac{3}{2}O_{2}\rightarrow Al_{2}O_{3}.....\Delta H_{1}=\frac{-3351.4}{2}kJ=-1675.7kJ$

$Cr_{2}O_{3}\rightarrow 2Cr+\frac{3}{2}O_{2}....\Delta H_{2}=\frac{2279.4}{2}kJ=1139.7kJ$

--------------------------------------------------------------------------------------------------------------

$2Al+Cr_{2}O_{3}\rightarrow 2Cr+Al_{2}O_{3}....\Delta H=\Delta H_{1}+\Delta H_{2}=-1675.7+1139.7kJ=-536kJ$

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3 years ago

Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...

g100num [7]

Answer:

O₂; KCl; 33.3

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

2KCl + 3O₂ ⟶ 2KClO₃

n/mol: 100.0 100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

$\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}$

(ii) From O₂

$\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}$

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

$\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}$

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

3 0

3 years ago