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Taya2010 [7]
2 years ago
5

A constant current of 0.350 A is passed through an electrolytic cell containing molten CrCl₂ for 21.7 h. What mass of Cr(s) is p

roduced? The molar mass of chromium is 52.0 g/mol. Provide your answer rounded to 3 significant digits.​
Chemistry
1 answer:
zhuklara [117]2 years ago
4 0

An electrochemical cell can generate or use electrical energy. The mass of solid chromium that will be deposited on the electrochemical plate is 7.17 gm.

<h3>What is current?</h3>

Current in an electrochemical cell is the ratio of the quantity of electricity in columns and time in seconds.

Given,

Current (I) = 0.350 A

Time = 21.7 hours

Molar mass of chromium = 52.0 g/mol

First time is converted into seconds:

1 hour = 3600 seconds

21.7 hours = 76020 seconds

The quantity of electricity flowing in the electrochemical solution is calculated as:

\begin{aligned} \rm Q & = \rm It\\\\& = 0.350 \times 76020 \\\\& = 26607\;\rm C \end{aligned}

Electricity required for depositing 1 mole or 52.0 g chromium is calculated as:

In electrochemical solution, chromium chloride is dissociated as:

\rm CrCl_{2} \rightarrow Cr^{2+} + 2 Cl^{-} \\\\\rm Cr^{2+} +2 e^{-} \rightarrow Cr

Two moles of electrons are needed to deposit 52.0 g of chromium.

If, 1 electron = 96500 C

Then, 2 electron = 193000 C

The mass of chromium deposited is calculated as:

193000 C = 52 g chromium

So, 26607 C = \dfrac{26607 \times 52}{193000} = 7.17 \;\rm gm

Therefore, 7.17 gm of chromium is produced.

Learn more about an electrochemical cell here:

brainly.com/question/20355190

#SPJ1

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Need help on last 3 questions
Olin [163]

Answer:

Explanation:

1)

Given data:

Initial volume of balloon = 0.8 L

Initial temperature = 12°C ( 12+273= 285 K)

Final temperature = 300°C (300+273 = 573 K)

Final volume = ?

Solution:

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 0.8 L .573 K / 285 K

V₂ = 458.4 L / 285

V₂ = 1.61 L

2)

Initial pressure = 204 kpa

Initial temperature = 29°C ( 29 + 273 = 302 K)

Final temperature = ?

Final pressure = 300 kpa

Solution:

P₁/T₁ = P₂/T₂

T₂ = T₁P₂/P₁  

T₂ = 302 K . 300 kpa / 204 kpa

T₂ = 90600 K/ 204

T₂ = 444.12 K

3)

Given data:

Initial volume = 14 L

Initial pressure = 2.1 atm

Initial temperature = 100 K

Final temperature = 450 K

Final volume = ?

Final pressure = 1.2 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm  

V₂ = 13230 L / 120

V₂ = 110.25 L

5 0
3 years ago
2NaCl + F2 yields 2NaF + Cl2
bearhunter [10]
Study your experiment setup.<span> In 30 minutes, how will the air temperature in the bottles compare?</span><span> What do you predict will happen to the ice in each bottle?</span>
7 0
3 years ago
5. A box with a volume of 22.4 L contains 1.0 mol of nitrogen and 2.0 mol of hydrogen at 0°C. Which of the following statements
love history [14]

B. The partial pressure of N2 is 101 kPa

<h3>Further explanation</h3>

Given

volume = 22.4 L

1.0 mol of nitrogen and 2.0 mol of hydrogen at 0°C

Required

Total pressure and partial pressure

Solution

Ideal gas law :

PV = nRT

n total = 3 mol

T = O °C + 273 = 273 K

P = nRT/V

P = 3 x 0.08205 x 273 / 22.4

P total = 3 atm = 303,975 kPa

P Nitrogen = 1/3 x 303.975 = 101.325 kPa

P Hydrogen = 2/3 x 303.975 = 202.65 kPa

7 0
3 years ago
Consider the reaction of gaseous hydrogen with gaseous oxygen to produce gaseous water. Given that the first picture represents
Bogdan [553]

The question is incomplete. There's missing the image, which is shown below.

Answer:

Volume of O₂ = 6 L, volume of mixture: 18 L, volume of H₂O = 12 L, molecule volume of H₂O = 0.667 molecule/L

Explanation:

The reaction between hydrogen gas and oxygen gas to form water is:

2H₂(g) + O₂(g) → 2H₂O(g)

So, for 1 mol of O₂ is necessary 2 moles of H₂ form 2 moles of H₂O. As the images below there's 8 molecules of H₂, 4 molecules of O₂, 12 molecules in the mixture, and 8 molecules of H₂O. Thus, there are stoichiometric values.

All the images are at the same temperature and pressure, so, by the ideal gas law:

PV= nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The number of moles and molecules are related, so let's substitute it in the equation. For the H₂:

P*12 = 8*RT

RT/P = 12/8 = 1.5

Thus, for O₂:

PV= nRT

V = n*(RT/P)

V = 4*1.5 = 6 L

For the mixture:

V = 12*1.5 = 18 L

For H₂O:

V = 8*1.5 = 12 L

The molecule volume is the number of molecules divided by the volume they occupy, thus for water: 8/12 = 0.667 molecules/L

6 0
3 years ago
My teacher is grading this soon can someone help me ASAP!
Stolb23 [73]

10. You demonstrated the difference in density of the two objects. It is a physical property.

11. First calculate the density for all of them: density = mass/volume

Density:

A. 5/6 g/ml

B. 10/9 g/ml

C. 15/16 g/ml

D. 20/10 g/ml

If the density of the substance is higher than the density of the substance it is put in, then it will sink. So substances B and D will sink in water, as their densities are higher than 1 g/ml.

12. Ammonia weighs less than water does-- for example, the weight of 8 gallons of ammonia will be equivalent to the weight of 5 gallons of water.

Hope this helped!

3 0
3 years ago
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