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anzhelika [568]
2 years ago
15

HELP ASAP!!!!!!!!

Chemistry
1 answer:
Irina-Kira [14]2 years ago
4 0
Answer is: <span>c. 80 g of Br2.
</span>1) n(Li) = 14 g ÷ 7 g/mol
n(Li) = 2 mol.
N(Li) = 2 mol · 6·10²³ 1/mol.
N(Li) = 1,2·10²⁴.
2) n(He) = 4 g ÷ 4 g/mol.
n(He) = 1 mol.
N(He) = 1 mol · 6·10²³ 1/mol.
N(He) = 6·10²³.
3) n(Br₂) = 80 g ÷ 160 g/mol.
n(Br₂) = 0,5 mol.
N(Br₂) = 0,5 mol · 6·10²³ 1/mol.
N(Br₂) = 3·10²³.
4) n(H₂) = 4 g ÷ 2 g/mol = 2 mol.
N(H₂) = 1,2·10²⁴.
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Answer:

Explanation:

As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)

ΔG = ΔG° + RTInQ

Q = 1

ΔG = ΔG°

ΔG = =nFE°

n=no of electrons transfered.

E° = 1.1v

ΔG° = -2 * 96500 * 1.10

= -212300J

ΔG° =-212.3kJ/mol

<h3>Therefore, the ΔG° = -212.3kJ/mol</h3>
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3 years ago
Is the electrolysis of water an oxidation-reduction reaction??
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Yes

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Check the explanation

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The answer to the question above is:

1) nonspontaneous because deltaG is positive

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Which type of chemical reaction takes place in an electrolytic cell
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2 years ago
The student ran out of time and did not do the second heating. Explain how this error would affect the calculation for the numbe
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The error caused will lower the number of moles when he did not do the second heating.This question is related to the empirical formula.

<h3>What is Empirical formula ?</h3>

A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

The empirical formula is the simplest ratio of atoms in a given compound.

we know that heating is used to eliminate most of the water molecules which are attached to the compound in the question, it's given that the second heating was not performed by the students.

This means that there are still some water molecules attached to the compound. This will imply that higher mass will be calculated for the compound because the water molecules are also considered.

So now, if we look at the calculation for the number of moles of water in this, we can say that this will be lower than the actual amount.

This is because there are some unaccounted moles of water which are still attached to the compound and these will be accounted in the compound mass rather than the waters. Therefore, the number of moles will be lower.

Hence, The error caused will lower the number of moles when he did not do the second heating.

Learn more about Empirical formula here ;

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