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Lynna [10]
3 years ago
11

How would you expect the appearance of a rock high in iron and magnesium to differ from a rock with very little iron and magnesi

um?
Chemistry
2 answers:
kondor19780726 [428]3 years ago
8 0

Answer:

As explained below.

Explanation:

  • The rocks that have a higher content of magnesium and iron content are found in igneous rocks composition that is mafic and felsic in nature have a higher concentration of silica content also and are highly intrusive n nature are highly explosive and have a more acidic base.
  • These rocks like the magmas of 750 to 950 degrees temperature are dark in color and higher pressure and others that don't have this high content are characterized as sedimentary or metamorphic in nature.  
  • <u>These have less density and light color with layered structures present in them.</u>
xenn [34]3 years ago
5 0
High iron and magnesium rocks have a high percentage of dark-colored (mafic) minerals.

So the high iron and magnesium would be darker than the low iron and magnesium.
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\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}

\text{and in the liquid form it is easily transported. An industrial chemist studying this}

\text{reaction fills a} \ \mathbf{100 \  L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \  \text{of oxygen gas, }

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Explanation:

From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

At equilibrium;

The water vapor = \dfrac{2.6 \ mol}{100 \ L} = 6x

x = \dfrac{2.6}{100} \times \dfrac{1}{6}

x = 0.00433

\text{equilibrium constant}  ({k_c}) =  \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }

\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\

Replacing the value of x, we have:

K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}

K_c = \mathbf{5.5 \times 10^{-8} \ to  \ 2 \ significant \ figures}

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