How will the candle contributes to the pressure and temperature of gases inside the glass?

1 answer:

7 0

How will the candle contributes to the pressure and temperature of gases inside the glass?

The candle contributes to the pressure and temperature of the gases inside the glass because of the increasing temperature the lit of the candle disposes into the gas molecules inside and as these gas molecules rise in temperature they become unsettling and since heat makes gas molecules robust the area becomes smaller for them to be ecstatic and as the dynamics is explained p=f/a the pressure increases.

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A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate

Zarrin [17]

Answer: Empirical formula is $C_2H_5O$

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

Step 1 : Converting each of the given masses into their moles by dividing them by Molar masses.

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = $\frac{330g}{12g/mol}=27.5moles$

Moles of Hydrogen = $\frac{69.5g}{1g/mol}=69.5moles$

Moles of Oxygen = $\frac{220.2g}{16g/mol}=13.76moles$

Step 2: Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

$\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2$

$\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5$

$\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1$

Step 3: Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = $C_2H_5O$

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