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3 years ago

Convert 25 gram of carbon dioxide to mold

Chemistry

1 answer:

Rus_ich [418]3 years ago

7 0

The atomic weight of carbon dioxide is about 12.011 + 15.999 + 15.999 = 44.009.

44.009 grams of CO2 = 1 mole of CO2 CO2 = Carbon dioxide

25 ÷ 44.009 ≈ 0.568

About 0.568 moles of CO2

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Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is comb

Bogdan [553]

Answer: The molecular formula for the menthol is $C_{10}H_{20}O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen 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 $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

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

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.0129g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles$

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 0.00066 moles.

For Carbon = $\frac{0.0064}{0.00066}=9.69\approx 10$

For Hydrogen = $\frac{0.0129}{0.00064}=19.54\approx 20$

For Oxygen = $\frac{0.00066}{0.00066}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is $C_{10}H_{20}O_1=C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156.27g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Thus, the molecular formula for the menthol is $C_{10}H_{20}O$

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3 years ago

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The mass of a nucleus is 0.042 amu less than the sum of the masses of 3 protons and 4 neutrons. The binding energy per nucleon i

Leni [432]

The answer is 10 mev. hope this helps

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3 years ago

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How many micrograms (ug) are in 3.4 x 10^-5 ounces (oz)

marysya [2.9K]

Answer:

964ug

Explanation:

The problem here involves converting from one unit to another.

We are to convert from ounces to micrograms.

1ug = 1 x 10⁻⁶g

1oz = 28.35g

So we first convert to grams from oz then take to ug:

Solving:

1oz = 28.35g

3.4 x 10⁻⁵oz will then give 3.4 x 10⁻⁵ x 28.35 = 9.64 x 10⁻⁴g

So;

1 x 10⁻⁶g = 1ug

9.64 x 10⁻⁴g will give $\frac{9.64 x 10^{-4} }{1 x 10^{-6} }$ = 9.64 x 10²ug or 964ug

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2 years ago

What is the independent variable, and the dependent variable in this?

Doss [256]

Answer:

The salt and sugar as well as the baby power and granulated sugar are the independent variables.

The dependent variable is how it reacts (will it dissolve or not)

Explanation:

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Please help ASAP. No one could not help me with this question.

Tanya [424]

Answer:

sorry

Explanation:

I don't know the answer this is really confusing but I am really sorry you have to do this.

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2 years ago