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3 years ago

Calculate the time required for a constant current of 0.868 A to deposit 0.387 g of Tl(III) as Tl(s) on a cathode.

Chemistry

1 answer:

dimaraw [331]3 years ago

6 0

Answer:

632.7 secs

Explanation:

Equation of the reaction:

Tl^3+ + 3e ------> Tl

If 1F = 96500 C

Q= It

Where;

I= current

t= time

3× 96500 C deposited 204 g of Tl

0.868 × t deposited 0.387 g of Tl

3 × 96500 × 0.387 = 204 × 0.868 × t

t= 3 × 96500 × 0.387/ 204 × 0.868

t= 632.7 secs

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Answer:

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A pair of students found the temperature of 100. g of water to be 26.5°C. They then dissolved 8.89 g of AgNO3 in the water. When

bekas [8.4K]

Explanation:

Initial temperature of the water = $T_i=26.5^oC$

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1 mole of silver nitrate gives 1 mole of silver ion i.e. cation.

Then 0.05229 moles of silver nitrate will give:

$\frac{1}{1}\times 0.05229 mol=0.05229 mol$ silver ions.

0.05229 moles of cations are produced.

1 mole of silver nitrate gives 1 mole of nitrate ion i.e. anion .

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3 years ago

Boyle's Law expresses the pressure-volume

Colt1911 [192]

Answer:

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Explanation:

Data:

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Calculation:

$\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{1.00 atm} \times \text{350. L} & = & p_{2} \times\text{2.00 L}\\\text{350. atm} & = & 2.00p_{2}\\p_{2} & = & \dfrac{\text{350. atm}}{2.00}\\\\& = &\textbf{175. atm}\\\end{array}\\\text{It would take a pressure of $\large \boxed{\textbf{175. atm}}$ to compress the gas.}$

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3 years ago

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geniusboy [140]

Answer:

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3 years ago

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Answer:

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