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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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A 3.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of
andrew-mc [135]

Answer: The molecular formula for the given organic compound X is C_6H_{8}O_7

Explanation:

We are given:

Mass of CO_2=4.13g

Mass of H_2O=1.13g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 4.13 g of carbon dioxide, =\frac{12}{44}\times 4.13=1.13g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 1.13 g of water, \frac{2}{18}\times 1.13=0.125g of hydrogen will be contained.

Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.13g}{12g/mole}=0.094moles

Moles of Hydrogen =\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.125g}{1g/mole}=0.125moles

Moles of Oxygen =\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.75g}{16g/mole}=0.109moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles

For Carbon = \frac{0.094}{0.094}=1

For Hydrogen = \frac{0.125}{0.094}=1.33

For Oxygen = \frac{0.109}{0.094}=1.16

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1: 1.33: 1.16

Converting them into whole number ratios by multiplying by 6:

The ratio of C : H : O = 6: 8: 7

Hence, the empirical formula for the given compound is C_6H_8O_7

Empirical mass = 6\times 12+8\times 1+7\times 16=192g

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

Putting values in above equation, we get:

n=\frac{192g/mol}{192g/mol}=1

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_6H_8O_7\times 1=C_6H_{8}O_7

Thus molecular formula for the given organic compound X is C_6H_{8}O_7

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