Explanation:
The given data is as follows.
Current (I) = 3.50 amp, Mass deposited = 100.0 g
Molar mass of Cr = 52 g
It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.
Therefore, 100 g of Cr will be deposited by "z" grams of electricity.
z =
= 185576.9 C
As we know that, Q = I × t
Hence, putting the given values into the above equation as follows.
Q = I × t
185576.9 C =
t = 53021.9 sec
Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.
Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
The answer is powder because if it was a small crystal it the molecules are tightly compact same with the small cube but there less compact, powder is loose and more spread out and easier to mix so it would react the fastest
Answer:
The cathode is the electrode where the reduction takes place.
Explanation:
Cell has three components:
an electrolyte and two electrodes which is a cathode and an anode.
Electrolyte is usually solution of the water or other solvents in which the ions are dissolved.
<u>In electrolytic cell:</u>
Negatively charged electrode is the cathode where the process of reduction takes place.
Positively charged electrode is the anode where the process of oxidation takes place.
<u>In galvanic cell:</u>
Positively charged electrode is the cathode where the process of reduction takes place.
Negatively charged electrode is the anode where the process of oxidation takes place.
<u>So, the correct answer is - The cathode is the electrode where the reduction takes place.</u>
1cm^3 = 1L would be the correct answer. One cubic centimeter equals .001 liter, so this equality above is not correct.
Please let me know if you have any questions! :)