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Setler [38]
2 years ago
11

Determine the number of moles in 3.13 × 10²³ formula units of MgCl₂.

Chemistry
1 answer:
love history [14]2 years ago
3 0

Answer:

<h2>0.52 moles </h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{3.13 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{3.13}{6.02}  \\  = 0.519933

We have the final answer as

<h3>0.52 moles</h3>

Hope this helps you

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5. What is the molar mass of ethanol (C2H5OH)?
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<h2>D) 46.07 g/mol</h2>

The molar mass is given by the sum of the atomic masses of the component elements of the substance.

In our case, the substance is ethanol or C₂H₅OH.

A_C=12.011

A_H=1.0079

A_O=15.9994

\mathcal M_{C_2H_5OH} = 2 \times A_C + 6 \times A_H + A_O = 2 \times 12.011 + 6 \times 1.00079+ 15.999 4= 24.022 + 6.0474 + 15.9994= 46.069≈ \boxed{46.07 \dfrac{g}{mol}}

5 0
2 years ago
A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513
igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is:  N₂O₄(g)  ⇆  2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

              2NO₂ (g)   ⇆   N₂O₄(g)      

Initially   2.20 mol              -

React          x                      x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq       (2.20-x) / 3.50L     (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) /  [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0  → Quadratic formula

a = 0.2933 ;  b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50  = 0.0971 M

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Explanation:

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