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3 years ago

HELP WHAT IS THE ANSWER TO ANY OF THESE !!!

Chemistry

2 answers:

photoshop1234 [79]3 years ago

8 0

Number 4 is
-Oxidation occurs at the anode, while reduction occurs at the cathode. Recharging a battery involves the conversion of electrical energy to chemical energy. During recharging, there is movement of electrons from an external power source to the anode, and on the other side electrons are removed from the cathode.

maks197457 [2]3 years ago

7 0

Answer:

3. They stop producing because they eventually run out of the anode metal so the reaction can not go on.

4. They reverse the flow of electrons when the voltage is above what it needs to go normally.

Explanation:

3. as the reaction occurs the cathode is gaining electrodes as the anode loses them so it gets to a point where the anode and shrunk down so much that it can't give off anymore electrons and that is when the battery stops.

4. lets say you have a cell with a voltage of 1.05 then you would need a voltage higher than that to make it go in reverse and recharge itself.

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Which of the following is a valid conversion factor? (3 points)

Natali5045456 [20]

1,000 mL/ 1 L
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zysi [14]

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3 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

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4 years ago

If a person had the values for an objects density and volume, what value can be calculated

zavuch27 [327]

Hello there! With the values of density and volume, you would be able to find the object's mass.

Density is found by dividing the mass by the volume, so you could place in the values of the density and the volume to get the mass.

For example:

500 = mass/10

The 500 being density and 100 being volume. You would use simple math rules and multiply 10 by 500, and you'd get 5000, therefore using the density and volume values and giving you the mass.

I hope I could help you and have a great day!

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3 years ago

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What is the mass of one mole of nitrogen atoms?

OLEGan [10]

Answer:

28.0

Explanation:

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