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3 years ago

During electrolysis, electrons flow from

Chemistry

2 answers:

mixer [17]3 years ago

8 0

Answer:

Explanation:

Electrolysis is defined as a process that uses electricity to propel a non-spontaneous chemical reaction. This technique is commercially used to separate elements from naturally occurring materials.

zlopas [31]3 years ago

6 0

A) an anode to a cathode.

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How many atoms are in potassium iodide,ki

timurjin [86]

Clap your hands each part word you eat and you will get your answer.

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3 years ago

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Roman mixes 12 liters of 8% acid solution with a 20% acid solution, which results in a 16% acid solution. Find the number of lit

inysia [295]

Let the 8% solution be A, the 20% solution be B and the final solution be C.

C = A + B
C = 12 + B

0.16C = 0.08(12) + 0.2(B)
0.16(12 + B) = 0.96 + 0.2B
0.96 = 0.04B
B = 24 Liters

C = 12 + 24
C = 36 Liters

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4 years ago

Which example is the site of oxidation when the dry cell is operating?

geniusboy [140]

Answer:

solid zinc layer

Explanation:

I took the test :)

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3 years ago

If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500

ratelena [41]

Answer: The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

Explanation:

We are given:

Initial concentration of sulfur dioxide = 0.500 M

Initial concentration of nitrogen dioxide = 0.500 M

Initial concentration of sulfur trioxide = 0.00500 M

Initial concentration of nitrogen monoxide = 0.00500 M

The chemical reaction follows:

$SO_2+NO_2\rightleftharpoons SO_3+NO$

Initial: 0.500 0.500 0.005 0.005

At eqllm: 0.500-x 0.500-x 0.005+x 0.005+x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}$

We are given:

$K_c=2.50$

Putting values in above equation, we get:

$2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37$

Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration

So, equilibrium concentration of sulfur dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of nitrogen dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of sulfur trioxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Equilibrium concentration of nitrogen monoxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

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3 years ago

Which of the following bonds is polar? a) F-F b) O-H c) O-O d) H-H

SSSSS [86.1K]

D) H-H
Hope this helps.

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3 years ago