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Marta_Voda [28]
3 years ago
8

A student claims that if she wanted to make a solution quickly, she should use small pellets instead of powder along with heatin

g and stirring. Do you agree or disagree with the student's claim? Answer using CER format. PLS HELP I’m a little confused and don’t know what to put.
Chemistry
1 answer:
torisob [31]3 years ago
4 0

Answer:

<em>I disagree with the student's claim</em>

<em></em>

Explanation:

<em>From basic chemistry, the rate of chemical reaction increases with an increased surface area.</em>

<em>This simply means that a powdered substance will react faster than a small pellet of the same substance.</em>

Let us break this down further. The pellet has a smaller exposed surface per grain to the solvent, here is how. Imagine the powdered substance bunched up together to make the small pellet, grains at the center of the pellet is blocked by other grains around it, and will take time before it comes in contact with the solvent. For the powdered grains, each grain has a surface independently exposed to the solvent and can easily come in contact with the solvent, taking less time to react with the solvent.

Technically, a powdered substance has more exposed surface area and logically will take a shorter time for the reaction to be complete.

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The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 2.001 g sample of B
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<u>Answer:</u> The empirical formula for the given compound is C_{15}H_{24}O_1

<u>Explanation:</u>

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

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

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

We are given:

Mass of CO_2=5.995g

Mass of H_2O=1.963g

Mass of sample = 2.001 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

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

So, in 5.995 g of carbon dioxide, \frac{12}{44}\times 5.995=1.635g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

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

So, in 1.963 g of water, \frac{2}{18}\times 1.963=0.218g of hydrogen will be contained.

Mass of oxygen in the compound = (2.001) - (1.635 + 0.218) = 0.148 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.635g}{12g/mole}=0.136moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.218g}{1g/mole}=0.218moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.148g}{16g/mole}=0.0092moles

  • <u>Step 2:</u> 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 0.0092 moles.

For Carbon = \frac{0.136}{0.0092}=14.78\approx 15

For Hydrogen = \frac{0.218}{0.0092}=23.69\approx 24

For Oxygen = \frac{0.0092}{0.0092}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 15 : 24 : 1

Hence, the empirical formula for the given compound is C_{15}H_{24}O_1

3 0
4 years ago
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