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Elodia [21]
3 years ago
10

Help!!!!?????????? plssssssss

Chemistry
2 answers:
Umnica [9.8K]3 years ago
5 0

Answer:

I think it is B

Explanation:

DIA [1.3K]3 years ago
4 0

Answer: it’s b)

Explanation: that’s the only difference that is listed

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Every day on his ride to school, Max sees some sedimentary rock. He starts to wonder: Could material from this sedimentary rock
Sloan [31]

Material from this sedimentary rock ever forms igneous rock, <u>Option D. Yes, if the sedimentary rock is moved below Earth’s outer layer and exposed to energy from Earth’s interior, it can melt into liquid rock and form </u><u>igneous rock.</u>

Sedimentary rocks are shaped from pre-existing rocks or pieces of soon-as-dwelling organisms. They form from deposits that collect on this planet's floor. Sedimentary rocks regularly have special layering or bedding.

Igneous rock, or magmatic rock, is one of the 3 primary rock kinds, the others being sedimentary and metamorphic. Igneous rock is shaped via the cooling and solidification of magma or lava. The magma may be derived from partial melts of existing rocks in either a planet's mantle or crust.

Learn more about igneous rocks here:-brainly.com/question/6533375

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<u>Disclaimer:- your question is incomplete, please see below for the complete question.</u>

Every day on his ride to school, Max sees some sedimentary rock. He starts to wonder: Could material from this sedimentary rock ever form igneous rock?

A. No, igneous rock can only form out of other igneous rocks. Sedimentary rock cannot change into igneous rock.

B. No, igneous rock forms under Earth’s outer layer due to energy from Earth’s interior, but the sedimentary rock is only at Earth’s surface.

C. Yes, if the sedimentary rock is exposed to energy from the sun at Earth’s surface for a long enough time, it can melt into liquid rock and form igneous rock.

D. Yes, if the sedimentary rock is moved below Earth’s outer layer and exposed to energy from Earth’s interior, it can melt into liquid rock and form igneous rock.

3 0
1 year ago
How many sigma and pi bonds, respectively, are in the molecule below? ch3ch2chchch3?
klio [65]
14 sigma bonds, 1 pi bond
8 0
3 years ago
15. Which of the following elements will not form a polar covalent bond with oxygen? A. Sodium B. Fluorine Oxygen D. Hydrogen
muminat

Answer:

C. oxygen

Explanation:

oxygen cannot for a polar covalent bond with another oxygen atom

3 0
3 years ago
In the reaction of iron water shown below what coefficient must be placed before the iron to balance the reaction fe+4h2o fe3o4
tiny-mole [99]
Answer is: three.
Unbalanced chemical reaction: Fe+4H₂O → Fe₃O₄ + 4H₂.
Balanced chemical reaction: 3Fe+4H₂O → Fe₃O₄ + 4H₂.
In balanced reaction, there is three iron atoms on the left and on the right, eight hydrogen atoms and four oxygen atoms on both side of chemical reaction.
4 0
3 years ago
Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graphite) + O2 --&
Darya [45]

Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ/mole

(2) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ/mole

(3) CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-726.4kJ/mole

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_1=-393.5kJ/mole

(2) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol

(3) CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g)     \Delta H_3=726.4kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole

6 0
3 years ago
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