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Marina86 [1]
3 years ago
14

Find the number of molecules in .815 mole of O2​

Chemistry
1 answer:
Simora [160]3 years ago
6 0

Answer:

4.9 × 10²³ molecules

Explanation:

Given data:

Number of molecules = ?

Number of moles of oxygen = 0.815 mol

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

0.815 mol  ×  6.022 × 10²³ molecules  / 1 mol

4.9 × 10²³ molecules

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What is the mole ratio that can convert from moles of A to moles of B?
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There is no direct way to convert from the mass of one substance to the mass of another. To convert from one mass (substance A) to another mass (substance B), you must convert the mass of A first to moles, then use the mole-to-mole conversion factor (B/A), then convert the mole amount of B back to grams of B.
8 0
3 years ago
A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.2KClO3 Right arrow. 2KCI + 3O2What is the perce
STALIN [3.7K]

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ →  2KCl  + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %

5 0
3 years ago
Hydrogen sulfide decomposes according to the following reaction: 2H2S(g) ⇋ 2H2(g) + S2(g) Kc=9.30x10-8 at 700.°C.If 0.45 mol of
Natalija [7]

Answer:

[H₂] = 1.61x10⁻³ M

Explanation:

2H₂S(g) ⇋ 2H₂(g) + S₂(g)

Kc = 9.30x10⁻⁸ = \frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}

First we <u>calculate the initial concentration</u>:

0.45 molH₂S / 3.0L = 0.15 M

The concentrations at equilibrium would be:

[H₂S] = 0.15 - 2x

[H₂] = 2x

[S₂] = x

We <u>put the data in the Kc expression and solve for x</u>:

\frac{(2x^2) * x}{(0.15-2x)^2}=9.30x10^{-8}

\frac{4x^3}{0.0225-4x^2}=9.30*10^{-8}

We make a simplification because x<<< 0.0225:

\frac{4x^3}{0.0225} =9.30*10^{-8}

x = 8.058x10⁻⁴

[H₂] = 2*x = 1.61x10⁻³ M

5 0
3 years ago
Read 2 more answers
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Got this off of quizlet I’m not 100% an expert but I tried to help
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