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2 years ago

Why is burning coal a major source of pollution? Respond in paragraph form. PLEASE HELP! GIVING BRAINLIEST!!

Chemistry

1 answer:

Alina [70]2 years ago

6 0

Answer:

Explanation:

Coal is chemically complex fuel. ... The sulfur in coal combines with oxygen to form sulfur dioxide, which can be a major source of air pollution if emitted in large enough quantities. Today, many of the effects of coal burning have been reduced significantly or eliminated.

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In a first order reaction 40% of reactant gets converted into product in 30 minutes. What time would it require to convert 75% i

e-lub [12.9K]

56 minutes and 15 seconds I believe

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2 years ago

The bonds in the compound MgSO4 can be described as

Butoxors [25]

C. Sulfur and oxygen (non metals) forms a covalent bond while the magnesium (a metal) will react with both non metals to form an ionic bond

7 0

2 years ago

Write down the dissolution equation for rubidium chromate dissolving in water. (Chromate is a polyatomic ion with the formula Cr

ycow [4]

Answer:

Four moles of the cation

Explanation:

2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)

Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.

The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.

This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.

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2 years ago

When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing po

SOVA2 [1]

The given question is incomplete. The complete question is as follows.

When 70.4 g of benzamide ( $C_{7}H_{7}NO$ ) are dissolved in 850 g of a certain mystery liquid X, the freezing point of the solution is $2.7^{o}C$ lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride ( $NH_{4}Cl$ ) are dissolved in the same mass of X, the freezing point of the solution is $9.9^{o}C$ lower than the freezing point of pure X.

Calculate the Van't Hoff factor for ammonium chloride in X.

Explanation:

First, we will calculate the moles of benzamide as follows.

Moles of benzamide = $\frac{mass}{\text{Molar mass of benzamide}}$

= $\frac{70.4 g}{121.14 g/mol}$

= 0.58 mol

Now, we will calculate the molality as follows.

Molality = $\frac{\text{moles of solute (benzamide)}}{\text{solvent mass in kg}}$

= $\frac{0.58 mol}{0.85 kg}$

= 0.6837

It is known that relation between change in temperature, Van't Hoff factor and molality is as follows.

dT = $i \times K_{f} \times m$ ,

where, dT = change in freezing point = $2.7^{o}C$

i = van't Hoff factor = 1 for non dissociable solutes

$K_{f}$ = freezing point constant of solvent

m = 0.6837

Therefore, putting the given values into the above formula as follows.

dT = $i \times K_{f} \times m$ ,

$2.7^{o}C = 1 \times K_{f} \times 0.6837 m$

$K_{f}$ = 3.949 C/m

Now, we use this $K_{f}$ value for calculating i for $NH_{4}Cl$

So, moles of ammonium chloride are calculated as follows.

Moles of $NH_{4}Cl = \frac{70.4 g}{53.491 g/mol}$

= 1.316 mol

Hence, calculate the molality as follows.

Molality = $\frac{1.316 mol}{0.85 kg}$

= 1.5484

It is given that value of change in temperature (dT) = $9.9^{o}C$ . Thus, calculate the value of Van't Hoff factor as follows.

dT = $i \times K_{f} \times m$

$9.9^{o}C = i \times 3.949 C/m \times 1.5484 m$

i = 1.62

Thus, we can conclude that the value of van't Hoff factor for ammonium chloride is 1.62.

5 0

2 years ago

The composition of dry air at sea level is 78.03% N2, 20.99% O2, and 0.033% CO2 by volume. (a) calculate the average molar mass

levacccp [35]

Answer:

the average molar mass of this air sample can be calculated as

addition of the product of the average molar weights of the component gases and their percentage compositions

1. Average Molar mass of Air = 0.7803 x 28 + 0.2099 x 32 + 0.00033 x 44 = 28.58g/mol

2. The partial pressures of N2, O2, and CO2 in atm.

From Ideal gas law, at stp, Volume of air V=22.4L/mol

PV =nRT

Since, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas

Total Pressure P=1atm

Partial Pressure p = mol fraction x P

Volume of N2 = 0.7803 x 22.4L = 17.47L, Partial Pressure = 0.7803atm

Volume of O2 = 0.2099 x 22.4L = 4.68L Partial Pressure = 0.209atm

Volume of CO2 = 0.00033 x 22.4L = 0.00739L, Partial Pressure = 0.033atm

7 0

2 years ago