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n200080 [17]
2 years ago
14

A compound is 48.6 % C, 8.2 % H, and 43.2 % O by mass. Calculate the empirical formula of this compound.

Chemistry
1 answer:
maria [59]2 years ago
6 0
Answer:
C2H3O
Explanation
:
C. H. O
48.6. 8.2. 43.2
r.a.m 12. 1. 16
no.of 48.4÷12. 8.2÷1. 43.2÷16
moles
4.05÷2.7. 8.2÷2.7. 2.7÷2.7
2. 3. 1
Therefore C2H3O
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Answer:

Now u have 48 g of O2. There. Fore mole=weight/M. W. Of oxygen. Therefor 3mole.

After that if we to multiply the avogadro number with it. So 3 *NA

Now u want only atom calculation then we have 2 molecule of oxygen then multiply it with 2 too.

So final claculation is =3*2*NA.

Explanation:

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What is the ground state electron configuration for magnesium?
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4 0
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3 0
3 years ago
What mass of zinc (molar mass 65.4 g moll) does it take to produce 0.50 mole of H2(g)?
Hatshy [7]

Answer:

32.7 g of Zn

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + 2HCl —> ZnCl₂ + H₂

From the balanced equation above,

1 mole of Zn reacted to produce 1 mole of H₂

Next, we shall determine the number of mole of Zn required to produce 0.5 mole of H₂. This can be obtained as follow:

From the balanced equation above,

1 mole of Zn reacted to produce 1 mole of H₂.

Therefore, 0.5 mole of Zn will also react to produce to 0.5 mole of H₂.

Thus, 0.5 mole of Zn is required.

Finally, we shall determine the mass of 0.5 mole of Zn. This can be obtained as follow:

Mole of Zn = 0.5 mole

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Mass of Zn =?

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Mass of Zn = 0.5 × 65.4

Mass of Zn = 32.7 g

Thus, 32.7 g of Zn is required to produce 0.5 mole of H₂.

4 0
2 years ago
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