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Debora [2.8K]
4 years ago
6

An electrically neutral atom bears what trait?

Chemistry
1 answer:
zhenek [66]4 years ago
4 0

C) It contains the same number of electrons and protons.

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Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o
Gnom [1K]

Answer: The empirical formula for the given compound is C_5H_7N

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

C_xH_yN_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of CO_2=21.363mg=21.363\times 10^3g=21363g

Mass of H_2O=6.125g=6.125\times 10^3g=6125g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, \frac{12}{44}\times 21363=5826.27g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, \frac{2}{18}\times 6125=680.55 of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = \frac{485.52}{97.73}=4.96\approx 5

For Hydrogen  = \frac{680.55}{97.73}=6.96\approx 7

For Nitrogen = \frac{97.73}{97.73}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is C_5H_7N_1=C_5H_7N

7 0
3 years ago
If there are 3.10 moles of O, how many moles of each of the compounds are present?
Burka [1]

Explanation:

The question pretty  much requires us to find the amount of moles of each compounds based on the number of moles of O given.

H2SO4

1 mol of H2SO4 contains 4 mol of O

x mol of H2SO4 would contain 3.10 mol of O

x = 3.10 * 1 / 4 = 0.775 mol of H2SO4

C2H4O2

1 mol of C2H4O2 contains 2 mol of O

x mol of C2H4O2 would contain 3.10 mol of O

x = 3.10 * 1 / 2 = 1.55 mol of C2H4O2

NaOH

1 mol of NaOH contains 1 mol of O

x mol of NaOH would contain 3.10 mol of O

x = 3.10 * 1 / 1 = 3.10 mol of NaOH

3 0
3 years ago
What happens to a glass of sugar solution when sugar is added to it?
galina1969 [7]

Answer:

The molarity of the solution increases.

Explanation:

Molarity is the measure of the concentration of the solute in the solution. In this case, the solvent is the sugar solution and the solute is the sugar.

If sugar is ADDED to the already sugary solution, then there would be more sugar. Therefore, the sugar (solute) would increase in number.

This means that the answer is the third choice: The molarity of the solution increases.

The answer would not be the first or second choice because there isn't anything in the question that implies water. It just says sugar solution.

The answer is not the last choice because the sugar concentration does not decrease after you have added more sugar to it. It increases.

7 0
4 years ago
Read 2 more answers
How many ways can you choose ten tadpoles from a pool of 20
Georgia [21]
Take half the amount
5 0
4 years ago
How many liters are in 5.42 mols of O2 gas?<br><br> (2 decimal places)
Debora [2.8K]

Explanation:

1 mol = 22.4 l

5.42 mol = 22.4 × 5.42 = 121.408

in two decimal place it is 121.41

7 0
3 years ago
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