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3 years ago

10/24/14

Chemistry

2 answers:

user100 [1]3 years ago

7 0

Answer:

D) Hydrogen and oxygen are being formed.

Explanation:

The compound water is breaking down into its elemental components and Hydrogen and oxygen are being formed

vlabodo [156]3 years ago

3 0

Answer:

d = Hydrogen and oxygen are being formed.

Explanation:

When current is passed through the water electrolysis take place.

Water is splitted into hydrogen and oxygen gas. Because of this formation bubbles are formed.

At cathode:

2H₂O + 2e⁻ → H₂ + 2OH⁻

Cathode is negatively charged and reduction take place on it.

At anode:

4OH⁻ → O₂ + 2H₂O + 4e⁻

Anode is positively charged and oxidation take place on it.

The over all reaction can be written as:

Chemical equation:

2H₂O (electricity) → 2H₂ + O₂

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Calculate the number of molecules in 100. g of O2 *

Sindrei [870]

I am pretty sure the answer is 6.25 molecules
See picture for explanation

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3 years ago

Nitrates are _____.

horrorfan [7]

Answer:

Nitrates are oxidising agents

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3 years ago

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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

Read 2 more answers

(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of

Fudgin [204]

Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

$P x V = k$