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Georgia [21]
3 years ago
9

How many atoms are present in a 1.0 mole sample of potassium perchlorate (KClO3)?

Chemistry
2 answers:
alekssr [168]3 years ago
5 0
<span>Okay, a mole of potassium perchlorate contains 6.02x1023 formula units of potassium perchlorate, but you're asking about individual atoms. So, let's look at the formula: KClO3. That's 1 potassium, 1 chlorine, and 3 oxygens, for a total of 5 atoms per formula unit. Now, multiple 5 by Avogadro's number above, to get 30.1x1023, which simplifies to 3.01x1024 atoms.</span>
andrew11 [14]3 years ago
3 0

<u>Answer:</u> The number of atoms in 1.0 moles of potassium perchlorate is 3.011\times 10^{24}

<u>Explanation:</u>

We are given:

Moles of potassium perchlorate = 1.0 moles

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

1 mole of KClO_3 contains 1 potassium atom, 1 chlorine atoms and 3 oxygen atoms.

Number of atoms in KClO_3 = 5

So, 1.0 moles of potassium perchlorate will contain = (1.0\times 6.022\times 10^{23})=3.011\times 10^{24} number of atoms.

Hence, the number of atoms in 1.0 moles of potassium perchlorate is 3.011\times 10^{24}

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In the equation CH4 + 2O2 --&gt; 2H2O + CO2 What is the mass of CO2 produced when 35g of O2 reacts?
Anestetic [448]

Answer:

24.06 g of CO₂

Explanation:

The balanced equation for the reaction is given below:

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Next, we shall determine the mass of O₂ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of O₂ = 2 × 16 = 32 g/mol

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Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Finally, we shall determine the mass of CO₂ produced by the reaction of 35 g of O₂. This can be obtained as follow:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Therefore, 35 g of O₂ will react to produce = (35 × 44)/64 = 24.06 g of CO₂.

Thus, 24.06 g of CO₂ were produced from the reaction.

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when we have the balanced equation of this half cell :

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