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Tom [10]
4 years ago
8

If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?

Chemistry
2 answers:
Arada [10]4 years ago
8 0

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

 

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

 

Answer:

<span>0.2 moles O</span>

Naya [18.7K]4 years ago
8 0

Answer : The moles of oxygen present in the sample is, 0.2 moles

Explanation : Given,

Moles of carbon = 0.200 mole

Moles of hydrogen = 0.400 mole

Mass of unknown compound = 6.00 g

Molar mass of carbon = 12 g/mole

Molar mass of hydrogen = 1 g/mole

Molar mass of oxygen = 16 g/mole

First we have to calculate the mass of carbon and hydrogen.

\text{Mass of carbon}=\text{Moles of carbon}\times \text{Molar mass of carbon}=(0.200mole)\times (12g/mole)=2.4g

\text{Mass of hydrogen}=\text{Moles of hydrogen}\times \text{Molar mass of hydrogen}=(0.400mole)\times (1g/mole)=0.4g

Now we have to calculate the mass of oxygen.

Total mass of unknown compound = Mass of carbon + Mass of hydrogen + Mass of oxygen

6.00 = 2.4 + 0.4 + Mass of oxygen

Mass of oxygen = 3.2 grams

Now we have tom calculate the moles of oxygen.

\text{Moles of oxygen}=\frac{\text{Mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.2g}{16g/mole}=0.2moles

Therefore, the moles of oxygen present in the sample is, 0.2 moles

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Answer:

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Explanation:

A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

H = \frac{E}{M*(T2-T1)}

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

H = \frac{6640 J}{0.348 Kg*(43.6-24.4) C} =\frac{6640 J}{6.686 KgC}

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The aluminium could be the metal, its specific heat is similar to that found in this problem.

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Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

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ΔQ = Mass × Specific heat capacity × Temperature change

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Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of  ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

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