ILL GIVE A BRAINLY PLS HELP

1 answer:

3 0

Atoms are the unit of the molecule of the compound. The 3.01 x 10²⁴ atoms of oxygen are present in 5 moles of water and 3 moles of carbon dioxide in the sample.

<h3>What are atoms?</h3>

Atoms are the smallest fundamental unit of the compounds that can be given by Avogadro's number.

For calculating the oxygen atoms in 5 mole water:

If 1 mole = 6.02 × 10²³

Then, 5 moles = 5 × 6.02 × 10²³

Hence, 3.01 x 10²⁴ atoms of oxygen are present in 5 moles of water.

Moles of carbon dioxide in the sample is calculated as:

If 1 mole of carbon dioxide = 6.02 × 10²³ molecules

Then moles in 1.8 x 10²⁴ molecules will be,

(1.8 x 10²⁴ molecules) ÷ (6.02 × 10²³ molecules) = 3 moles

Hence, 3 moles of carbon dioxide is present in the sample.

Learn more about Avogadro's number here:

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This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:

<h3>Heating curves:</h3>

In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.

Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

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Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:

$Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\
\\
Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\
\\
Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\
\\
Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\
\\
Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ$