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mina [271]
3 years ago
10

A chemist prepares a solution of aluminum chloride (AlCl3) by measuring out 11 gr of aluminum chloride (AlCl3) into a 50 ml volu

metric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's aluminum chloride solution. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

1.6 M

Explanation:

To answer this question first we <u>calculate the molar mass of AlCl₃</u>, using the atomic weights of Al and Cl:

  • AlCl₃ ⇒ 27 g/mol + (35.45 g/mol)*3 = 133.35 g/mol

Now with the molar mass, and the mass weighed, we can <u>calculate mol AlCl₃</u>:

  • 11 g AlCl₃ ÷ 133.35 g/mol = 0.082 mol AlCl₃

Finally we <u>calculate concentration in mol/L</u>, keeping in mind that 50mL = 0.05 L:

  • 0.082 mol AlCl₃ / 0.05 L = 1.6 M
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There are many different types of preservatives like Benzoic acid, Calcium Sorbate, Erythorbic Acid, Potassium Nitrate and Sodium Benzoate. Some act like antioxidants used for slowing down spoilage like Ascorbyl Palmitate, Butylated Hydroxy anisole (BHA) and Butylated Hydroxytoluene (BHT

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2 years ago
How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten mgcl2 with an appl
ratelena [41]
First, we need to determine the half reaction of magnesium. It would be expressed as:

Mg2+ + 2e- = Mg

Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:

4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C

We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.

35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h
3 0
3 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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