What are the two kinds of electricity?

The three primary sources of air pollution are _____.

Kisachek [45]

There are four main types of air pollution sources:

mobile sources – such as cars, buses, planes, trucks, and trainsstationary sources – such as power plants, oil refineries, industrial facilities, and factoriesarea sources – such as agricultural areas, cities, and wood burning fireplacesnatural sources – such as wind-blown dust, wildfires, and volcanoes........hope i helped

7 0

3 years ago

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Suppose you have an Avogadro's number of nitrogen atoms. How many grams does this represent? Enter your answer to one decimal pl

shusha [124]

Answer:

2.3 x 10-23 g.

Explanation:

The mass of a single atom is the mass number, 14, is the mass in grams of one mole of carbon.

One mole of Nitrogen atom is 6.022 x 1023 atoms (Avogadro's number). This can then used to convert a nitogen atom to grams by the ratio:

mass of 1 atom / 1 atom = mass of a mole of atoms / 6.022 x 10^23 atoms.

mass of 1 atom = mass of a mole of atoms / 6.022 x 1023

mass of 1 N atom = 14 / 6.022 x 10^23 N atoms

mass of 1 N atom = 2.325 x 10^-23 g

The mass of a single Nitrogen atom is 2.325 x 10-23 g.

3 0

3 years ago

You have 15.42g of C2H6. How many moles of H2O can be made?

Amiraneli [1.4K]

Answer: The moles of water produced are 1.54 moles.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ethane = 15.42 g

Molar mass of ethane = 30.07 g/mol

Putting values in above equation, we get:

$\text{Moles of ethane}=\frac{15.42g}{30.07g/mol}=0.513mol$

The chemical equation for the combustion of ethane follows:

$2C_2H_6+5O_2\rightarrow 4CO2+6H_2O$

By Stoichiometry of the reaction:

2 moles of ethane produces 6 moles of water

So, 0.513 moles of ethane will produce = $\frac{6}{2}\times 0.513=1.54mol$ of water

Hence, the moles of water produced are 1.54 moles.

4 0

3 years ago

Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

scoray [572]

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion= $\frac{Volume}{Time}$

Effusion rate of oxygen gas after time t = $E=\frac{4.64 mL}{t}$

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = $E'=\frac{3.12 mL}{t}$

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$