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3 years ago

Identify 3 physical changes that could occur with a sheet of paper

Chemistry

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

1. Dimensions: The sheet can be cut into several smaller pieces.

2. Shape: the sheet can be formed into a paper ball or any other shape (using origami for example).

3. Color: You can use crayons to paint the sheet and change its color.

Explanation:

Physical changes do not change the chemical properties. They do not transform the compounds that form the substances. They do not break and or form chemical bonds.

When you change the form, the state (solid, liquid or gas), the color, you are just changing physical properties.

Only chemical changes change the compounds that form the substance. For instance, if you burn the sheet of paper, then you are causing a chemical change because the organic matter in the paper will react with oxygen forming CO₂ and water, but by cutting, folding, smashing, or coloring with crayons, the compounds in the sheet of paper do not change.

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Caproic acid, responsible for the odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this

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Answer: Empirical formula = C3H6O

C6H1202---Molecular formula

Explanation:

we first find the masses of C, H and O contained in the sample

mass of carbon

Mass of CO2 = 0.512g

molar mass of carbon = 12g/mol

Molar mass CO2 = 12+ (16X2)=44g/mol

Mass C = 12/44x 0.512 = 0.1396g of carbon

Mass Hydrogen

mass of H2O= 0.209g

Molar mass H2O = 18g/mol

Molar mass Hydrogen = 1.0079g x 2 = 2.0158g since Hydrogen gas is diatomic

Mass H = 2.0158/18 X 0.209 = 0.0234g Hydrogen

Mass of Oxygen in the sample = overall mass of Sample - mass of hydrogen and oxgen=

0.225 - ( 0.1396 + 0.0234) = 0.062g Oxygen

A) T o find Empirical formula

1ST STEP-- Divide through by each relative atomic mass:

C = 0.1396/12 = 0.01163

H = 0.0234/1.0079 = 0.0232

O = 0.0619/16 = 0.00387

2nd step Divide by smallest answer

C = 0.01163/0.00387 = 2.0998 = 3

H = 0.0232/0.00387 = 5.99 = 6

O = 0.00387/0.00387=1

Empirical formula = C3H6O

B)To find molecular formula

Given that molar mass of caproic acid as 116g,

we will use our empirical formulae to find molecular formulae with the equation

(C3H6O)n= 116g/mol

12x3 + 1x6 + 1 x16= 58g/mol

58n= 116g/mol

n = 116/58= 2

= (C3H6O)2= C(3X2) H(6X2) O(1X2)

C6H1202--- MOLECULAR FORMULA

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Compounds X and Y are both C6H13Cl compounds formed in the radical chlorination of 3-methylpentane. Both X and Y undergo base-pr

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Answer:

Y is a 3-chloro-3-methylpentane.

The structure is shown in the figure attached.

Explanation:

The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).

The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.

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Answer:

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