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miskamm [114]
3 years ago
7

Do cfcs break down ozone

Chemistry
1 answer:
frozen [14]3 years ago
8 0

Answer:

Once in the atmosphere, CFCs drift slowly upward to the stratosphere, where they are broken up. So true.

Hope this helps! (づ ̄3 ̄)づ╭❤~

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The powder could be acetaminophen, analgesic  having chemical formula C_{17}H_{21}NO_{4}

<h3>What is an empirical formula?</h3>

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.

We are given:

Percentage of C = 67.31 %

Percentage of H = 6.978 %

Percentage of N = 4.617 %

Percentage of O = 21.10 %

Let the mass of the compound be 100 g. So, the percentages given are taken as mass.

Mass of C = 67.31 g

Mass of H = 6.978 g

Mass of N = 4.617 g

Mass of O = 21.10 g

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

Step 1: Converting the given masses into moles.

Moles of carbon = \frac{mass}{molar \;mass}

Moles of carbon =\frac{67.31g}{12g/mole}

=5.60 moles

Moles of hydrogen = \frac{mass}{molar \;mass}

Moles of hydrogen = \frac{6.978 g}{1 g/mole}

=6.978 moles

Moles of nitrogen =\frac{mass}{molar \;mass}

Moles of nitrogen = \frac{4.617  g}{14 g/mole}

=0.329 moles

Moles of oxygen =\frac{mass}{molar \;mass}

Moles of oxygen =\frac{21.10 g}{16 g/mole}

=1.31 moles

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 0.329 moles.

We get the ratio of C : H : N : O = 17 : 21 : 1 : 4

The empirical formula for the given compound is C_{17}H_{21}NO_{4}.

Learn more about the empirical formula here:

brainly.com/question/14044066

#SPJ1

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