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Rina8888 [55]
4 years ago
6

Determine the percent composition of all elements in enstatite, MgSiO3, a mineral found in Earth's crust.

Chemistry
2 answers:
kodGreya [7K]4 years ago
8 0
<span>Magnesium:
Mg molar mass: 24.3050
You have one atom of Mg so, <span>24.211% (mass percent)

</span></span><span>Silicon:
Si molar mass: 28.0855
You have one atom of Si so, <span>27.977% (mass percent)

</span></span><span>Oxygen:
O molar mass: 15.9994
You have three atoms so, <span>47.812% (mass percent)</span></span>
rusak2 [61]4 years ago
5 0

Explanation:

Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100

Mass of magnesium atom = 24 g/mol

Mass of silicon atom = 28 g/mol

Mass of oxygen atom = 16 g/mol

Molar mass of MgSiO_3 =24 g/mol+28 g/mol+3 × 16 g/mol=100 g/mol

Composition of all elements in enstatite, MgSiO_3

Percentage of magnesium :

\frac{1\times 24 g/mol}{100 g/mol}\times 100=24\%

Percentage of silicon:

\frac{1\times 28 g/mol}{100 g/mol}\times 100=28\%

Percentage of oxygen:

\frac{3\times 16 g/mol}{100 g/mol}\times 100=48\%

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satela [25.4K]

Answer:

31395 J

Explanation:

Given data:

mass of water = 150 g

Initial temperature = 25 °C

Final temperature = 75 °C

Energy absorbed = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 75 °C - 25 °C

ΔT = 50 °C

now we will put the values in formula

q = m . c . ΔT

q = 150 g × 4.186 J/g.°C × 50 °C

q = 31395 J

so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .

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Consider the mechanism. Step 1: A+B↽−−⇀CA+B↽−−⇀C equilibrium Step 2: C+A⟶DC+A⟶D slow Overall: 2A+B⟶D2A+B⟶D Determine the rate la
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Explanation:

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\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]

2. Derive the rate law

Assume k₋₁ ≫ k₂.  

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In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is  

rate = k[A][B] where k = k₂K

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