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3 years ago

How many moles are in 123.0 grams of KClO4? (3 points)

Chemistry

1 answer:

iVinArrow [24]3 years ago

5 0

Answer:

0.8878 mol KClO4

Explanation:

Molar mass of KClO4 = 39.0983 + 35.453 + ( 15.9994 x 4 ) = 39.0983 + 35.453 + 63.9976 = 138.5489g/mol

but,

number of moles = mass in grams / molar mass where mass in grams = 123g, molar mass = 138.5489g/mol

number of moles = 123g / 138.5489g/mol = 0.8878mol

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What causes polar bonds

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Answer:

Bonds are polar when one element in a compound is more electronegative than the other. This creates a dipole in the molecule where one end of the molecule is partially positive and one end is partially negative

Explanation:

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3 years ago

The Law of _____states that substances combine in predictable proportions and that excess reactants remain unchanged.

RideAnS [48]

Answer: definite proportions.

Explanation:

1) The definite proportions law states that compounds will always have the same kind of atoms (elements) in the same mass proportion (ratios).

2) For example, a molecule of water will alwys have the same mass ratio of hydrogen atoms to oxygen atoms. That is what permits to obtain the chemical formula of the water molecule as H₂O.

The mass of the two hydrogen atoms will be in a fixed ratio respect to the mass of the oxygen atoms.

Then, if you have one reactant in less proportion than the other, respect to the ratio stated by the chemical formula of water, the former will react completely (it is the limiting reactant) with the corresponding (proportional) mass of the later. Then there will be an excess of the later reactant which will not react (will remain unchanged).

The reactants can only react in the proportion defined by the chemical formulas of the final products.

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3 years ago

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Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

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

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

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 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

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

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

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

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

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 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

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3 years ago

Help please!!!! I only have a couple mins

Kaylis [27]

Answer:

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Explanation:

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3 years ago

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What past or current environments allowed Nuclear Power to form?

Nimfa-mama [501]

Answer:

creation of radioactive wastes such as uranium mill tailings, spent (used) reactor fuel, and other radioactive wastes.

Explanation:

Nuclear energy produces radioactive waste

A major environmental concern related to nuclear power is the creation of radioactive wastes such as uranium mill tailings, spent (used) reactor fuel, and other radioactive wastes. These materials can remain radioactive and dangerous to human health for thousands of years.

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