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2 years ago

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- Equilibrium shifts for slightly soluble compounds The reaction for the formation of a saturated solution of silver bromide (Ag

Br) can be represented as AgBr(s)?Ag+(aq)+Br?(aq)
Consider that this reaction is an endothermic process and that the AgBr solution is saturated.

Classify the following changes based on whether they will favor the formation of reactant, whether they will favor the formation of products, or whether they will have no effect on the equilibrium.

1. Adding hydrobromic acid (HBr)

2. Lowering the concentration of bromide (Br-) ions

3. Heating the concentrations

4. Adding solid silver bromide (AgBr)

5. Removing silver (Ag+) ions

1 answer:

RideAnS [48]2 years ago

3 0

Answer:

1) Favor formation of reactant

2) Favor formation of product

3) Favor formation of product

4) Favor formation of product

5) Favor formation of product

Explanation:

The reaction is $AgBr(s)--->Ag^{+}(aq)+Br^{-}(aq)$

The reaction is endothermic

1. Adding hydrobromic acid (HBr)

It will increase the amount of bromide ion. Bromide ion is one of the product so it will favor formation of reactant.

2. Lowering the concentration of bromide (Br-) ions

As we are decreasing the the concentration of product it will favor formation of more products

3. Heating the concentrations

If we are heating the mixture, it will increase the temperature and it will favor endothermic reaction. Thus will favor formation of product.

4. Adding solid silver bromide (AgBr)

If we are adding silver bromide it means we are increasing the concentration of reactant, it will favor formation of products

5. Removing silver (Ag+) ions

Removing silver ions means we are decreasing the concentration of product thus it will favor formation of products.

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A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 417 pm. The diameter of the metal atom is:

enyata [817]

Answer:

b. 295 pm

Explanation:

To answer this question we need to use the equation of a face-centered cubic laticce:

Edge length = √8 R

<em>Where R is radius of the atom.</em>

<em />

Replacing:

417pm = √8 R

R = 147.4pm is the radius of the atom

As diameter = 2 radius.

Diameter of the metal atom is:

147.4pm* 2 =

295pm

Right answer is:

<h3>b. 295 pm </h3>

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What is a characteristic of all fuel cells?

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B. Electrical energy is produced from oxidation reactions.
I don't have an explanation for this though. Do you need one? I can probably look it up.

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Which compound has the lowest vapor pressure at 50°C?

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Answer : Option A) Ethanoic Acid

Explanation : Ethanoic acid has the lowest vapor pressure i.e 0.08 atm at the temperature of 50°C compared to the other given options.

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Ethanol has vapour pressure as 0.30 atm at 50°C

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3. According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density o

velikii [3]

Answer:

a) 11.64 M

b) 43 mL

c) 1.7 kg

Explanation:

a) Let's use a basis of the calculus of 1000 mL (1 L) of the concentrated solution. If the solution has 1.18 g/mL, it has:

1.18*1000 = 1180 g.

The mass of HCl will be then:

mHCl = 1180*0.36 = 424.8 g

The molar mass of HCl is 36.5 g/mol, so the number of moles is the mass divided by the molar mass:

nHCl = 424.8/36.5 = 11.64 mol

The molarity is the number of moles divided by the volume in L:

Molarity = 11.64 M

b) To prepare a solution by dilution of a concentrated one, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 the final solution. So:

11.64*V1 = 2.00*0.250

V1 = 0.0429 L ≅ 43 mL

c) The neutralization will happen by the equation:

HCl + NaHCO₃ → NaCl + CO₂ + H₂O

So, 1 mol of NaHCO₃ is needed to react with 1 mol of HCl. At 1.75 L, the number of moles of the acid is:

nHCl = 1.75*11.64 = 20.37 mol

The molar mass of NaHCO₃ is 84 g/mol so the mass needed is the molar mass multiplied by the number of moles:

m = 84*20.37 = 1,711.08 g

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2 years ago

Calculate the daily aluminum production of a 150,000 [A] aluminum cell that operates at a faradaic efficiency of 89%. The cell r

Gala2k [10]

Explanation:

It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.

$24 \times 3600 sec$

Hence, total charge passed daily is calculated as follows.

$150,000 \times 24 \times 3600 sec$

And, number of Faraday of charge is as follows.

$\frac{150,000 \times 24 \times 3600 sec}{96500}$

= 134300.52 F

The oxidation state of aluminium in $Al_{2}O_{3}$ is +3.

$Al^{3+} + 3e^{-} \rightarrow Al(s)$

So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.

Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.

$\frac{134300.52 F}{3} \times \frac{89}{100}$

= 39842.487 mol

or, = $3.9842 \times 10^{4} mol$

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

Mass in grams = $3.9842 \times 10^{4} mol \times 27$

= $107.57 \times 10^{4} g$

= 1075.7 kg/day

Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.

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