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2 years ago

Consider the following redox reaction: Ag⁺(aq) + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq). How many moles of electrons are exchanged?

Chemistry

1 answer:

Kisachek [45]2 years ago

5 0

Answer:

1 mol of electron is exchanged. The mol of electrons that is released by the iron, is gained by the silver.

Explanation:

We need to determine the half reactions:

Ag⁺ → Ag

These is the reduction reaction, where Silver decreases the oxidation state. It gained 1 mol of e⁻

Fe²⁺ → Fe³⁺

In this case, iron increases the oxidation state, from +2 to +3. It has released 1 mol of electrons. This is the oxidation reaction.

The complete redox is:

Ag⁺ + 1e⁻ → Ag

Fe²⁺ → Fe³⁺ + 1e⁻

Ag⁺(aq) + 1e⁻ + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq) + 1e⁻

The electrons will be cancelled.

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Select the correct answer. A student places a sample of white sodium bicarbonate powder in a test tube and heats it. The student

inessss [21]

Answer is: D. It is not sodium bicarbonate.

Balanced chemical reaction of heating sodium bicarbonate: 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O.

This is chemical change (chemical reaction), because new substances are formed (sodium carbonate, carbon(IV) oxide and water), the atoms are rearranged, so there is no sodium bicarbonate (NaHCO₃) in the test tube.

8 0

2 years ago

If excess caso4(s) is mixed with water at 25 ∘c to produce a saturated solution of caso4, what is the equilibrium concentration

Ray Of Light [21]

Answer:

The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M

Data Given:

Ksp of CaSO4 = 2.4 × 10∧-5

CaSO4 ⇔ Ca+2 + SO4∧-2

Solution:

Ksp = [Ca+2].[ SO4∧-2]

2.4 × 10∧-5 = [x].[x]= x²

x = 4.9×10∧-3 M

Result:

The conc. of Ca+2 ion is 4.9×10∧-3 M

3 0

2 years ago

Beer brewing begins with steeping grains in hot water, releasing the sugars inside. The sugar water is then heated to a boil and

user100 [1]

Answer:

The answers to the question are

a. 166.64 ° F

b. 217990.08 J/hour or 60.55 J/s = 60.55 watts

c. 13.C

Explanation:

a. To solve the question we list out the given variables thus

mass of grain = 16.5 lbs

Temperature of grain = 67 °F

Volume of hot water = 5 gals = ‪0.02273‬ m³

Equilibrium temperature of the mixture = 154 °F

Specific heat capacity of the grain = 0.44 times specific heat capacity of water

Therefore we have

Heat supplied by hot water = heat gained by mixture

Density of the water = 997 kg/m³ which gives

Therefore the mass of the water = (Density of the water) × (Volume of the water) = (997 kg/m³) × ‪(0.02273‬ m³) = 22.66181 kg

Therefore the heat supplied by the water =22.66 kg×1000 g/kg ×4.2 J/g°C×(Tₓ -‪67.78 °C) = ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×(67.78 -‪19.44)

= 95172 × (Tₓ -‪67.78 °C) =668205.7536 J

(Tₓ -‪67.78 °C) = 7.02 from where Tₓ = 74.80 °C = ‪166.64 ° F

The initial temperature (strike temperature) of the hot water = 74.80 °C = 166.64 ° F

b. Where the mixture lost two degrees we have

22.66 kg×1000 g/kg ×4.2 J/g°C×2 °C + ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×2 °C = 217990.08 J therefore the average energy lost per unit time = 217990.08 J/hour or 60.55 J/s

c. To find out how much it cost we have

Heat energy required to raise 5 gallons of water from 110 °F to 166.64 °F we have

22.66 kg×1000 g/kg ×4.2 J/g°C×(74.8 °C-‪43.33 °C) = 2994745.92 J

Energy lost during the heating = 10% = 299474.59 J

Total energy supplied 2994745.92 J + 299474.59 J = 3294220.5 J

Time for heating = 47 minutes, therefore rate of energy consumption = (3294220.5 J)/ (47×60) = 1168.163 Watt 1.168 kW

Cost of energy = 15.C per kilowatt-hour therefore 1.168 kW for 47 minutes will cost

1.168 kW ×47/60×15 = 13.C

therefore it cost 13.C to heat the 5 gallons of tap water initially at 110 ° F to the strike temperature 166.64 °F

6 0

2 years ago

Determine the percent composition by mass of a 100g salt solution which contains 20g salt

prisoha [69]

Answer:

Mass of solution=100g

mass of salt=20g

so; mass of solute=80g

percentage composition =(mass of salt/total

mass) ×100

= \frac{20}{100} \times 100 \\ = 20\%

glad to help you

hope it helps

8 0

1 year ago

Read 2 more answers

A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2

4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

$C_3H_7+5O_2\rightarrow 3CO_2+4H_2O$

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

$\frac{3}{1}\times 3.85 mol=11.55 mol$ of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

$\frac{4}{1}\times 3.85 mol=15.4 mol$ of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

$\frac{4}{5}\times 0.647 mol=0.5176 mol$ of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

$\frac{1}{5}\times 0.647 mol=0.1294 mol$ of propane.

0.1294 moles of propane are consumed.

5 0

2 years ago