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loris [4]
3 years ago
12

Why do copper ions go to the negative electrode

Chemistry
1 answer:
Snowcat [4.5K]3 years ago
8 0
Its because the copper ions have a positive charge, therefore, they are attracted to a negative electrode.
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1. This balanced equation represents a chemical reaction.
Dafna1 [17]

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combustion is the answer

5 0
2 years ago
If a dog has a mass of 16.1 kg, what is its mass in the following units? Use scientific notation in all of your answers.
nekit [7.7K]

Answer:

16.1\times 10^3\ \text{g}

16.1\times 10^{6}\ \text{mg}

16.1\times 10^{9}\ \mu\text{g}

Explanation:

Mass of dog = 16.1\ \text{kg}

1\ \text{kg}=10^{3}\ \text{g}

1\ \text{kg}=10^{6}\ \text{mg}

1\ \text{kg}=10^{9}\ \mu\text{g}

So,

16.1\ \text{kg}=16.1\times 10^3\ \text{g}

16.1\ \text{kg}=16.1\times 10^{6}\ \text{mg}

16.1\ \text{kg}=16.1\times 10^{9}\ \mu\text{g}

Mass of the dog in the other units is 16.1\times 10^3\ \text{g}, 16.1\times 10^{6}\ \text{mg} and 16.1\times 10^{9}\ \mu\text{g}.

3 0
2 years ago
In which of the following reactions does a decrease in the volume of the reaction vessel at constant
Black_prince [1.1K]

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the <u>volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.</u>

<u />

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

<u>Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby </u><em><u>favoring the formation of products.</u></em>

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

<u>Therefore, when the volume decreases, the equilibrium does not shift in any direction.</u>

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

<u>Therefore, when the volume decreases, the equilibrium does not shift in any direction.</u>

D) 2O₃(g) → 3O₂(g)

The number of moles of reactant is 2 and number of moles of product is 3.

<u>Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby </u><em><u>favoring the formation of reactants.</u></em>

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

<u>Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby </u><em><u>favoring the formation of reactants.</u></em>

6 0
3 years ago
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