1. Calculate the number of moles of carbon atoms present in 2.4g of ethanol (C2H6O).

2. What is the % Nitrogen by weight in 3.696 mol of ammonium sulfide.

3. How many atoms of mercury are present in 5.4 cubic cm of liquid mercury? The density of mercury is 13.55g/cc.

1. moles = grams/molar mass

2. (mass nitrogen/mass amm sulfide)*100 = ??
3. mass Hg = volume x density
Then convert to moles.
Then remember there are 6.022E23 atoms for every mole.

3.50 mol x 6.022 x 10^23

= 2.11 x 10^24 carbon atoms

1. To calculate the number of moles of carbon atoms in ethanol (C2H6O), we need to determine the molar mass of carbon. The molar mass of carbon (C) is approximately 12.01 g/mol.

First, we need to calculate the molar mass of ethanol:
C2H6O = (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 24.02 g/mol + 6.06 g/mol + 16.00 g/mol
= 46.08 g/mol

Next, we can use the molar mass of ethanol to calculate the number of moles of ethanol in 2.4 g:
Number of moles = Mass of substance (g) / Molar mass (g/mol)
Number of moles of ethanol = 2.4 g / 46.08 g/mol
≈ 0.052 moles of ethanol

Finally, since there are 2 carbon atoms in 1 mole of ethanol, we can calculate the number of moles of carbon atoms:
Number of moles of carbon atoms = 0.052 moles of ethanol * 2
≈ 0.104 moles of carbon atoms

2. To determine the percentage of nitrogen by weight in ammonium sulfide (NH4)2S, we need to calculate the molar mass of nitrogen (N) and the molar mass of ammonium sulfide.

The molar mass of nitrogen (N) is approximately 14.01 g/mol.

The molar mass of ammonium sulfide [(NH4)2S] can be calculated as follows:
[(2 * 14.01 g/mol) + (8 * 1.01 g/mol)] * 2 + 32.07 g/mol
= 28.02 g/mol + 16.08 g/mol + 32.07 g/mol
= 76.17 g/mol

Now we can calculate the weight percentage of nitrogen:
Weight percentage of nitrogen = (Molar mass of nitrogen / Molar mass of ammonium sulfide) * 100
= (14.01 g/mol / 76.17 g/mol) * 100
≈ 18.36%

Therefore, the % nitrogen by weight in 3.696 mol of ammonium sulfide is approximately 18.36%.

3. To determine the number of atoms of mercury in 5.4 cubic cm of liquid mercury, we first need to calculate the mass of mercury using its given density.

The density of mercury is 13.55 g/cc, which means that 1 cc (or 1 cubic cm) of liquid mercury weighs 13.55 grams.

The mass of 5.4 cubic cm of liquid mercury can be calculated as follows:
Mass = Density * Volume
= 13.55 g/cc * 5.4 cc
≈ 73.17 g

Next, we need to calculate the number of moles of mercury using its molar mass. The molar mass of mercury (Hg) is approximately 200.59 g/mol.

Number of moles = Mass (g) / Molar mass (g/mol)
Number of moles of mercury = 73.17 g / 200.59 g/mol
≈ 0.365 moles of mercury

Finally, we can calculate the number of atoms of mercury using Avogadro's number. Avogadro's number is approximately 6.022 × 10^23 atoms/mol.

Number of atoms = Number of moles * Avogadro's number
Number of atoms of mercury = 0.365 moles of mercury * 6.022 × 10^23 atoms/mol
≈ 2.20 × 10^23 atoms of mercury

Therefore, there are approximately 2.20 × 10^23 atoms of mercury in 5.4 cubic cm of liquid mercury.

1. To calculate the number of moles of carbon atoms in ethanol, we need to determine the molar mass of carbon. Therefore, we can use the molar mass of carbon to convert the given mass of ethanol to moles of carbon atoms.

First, we need to determine the molar mass of ethanol (C2H6O):
Molar mass = (2 * Atomic mass of Carbon) + (6 * Atomic mass of Hydrogen) + (1 * Atomic mass of Oxygen)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 24.02 g/mol + 6.06 g/mol + 16.00 g/mol
= 46.08 g/mol

Next, we can use the molar mass of ethanol to calculate the number of moles of ethanol:
Number of moles = Given mass / molar mass
Number of moles = 2.4 g / 46.08 g/mol ≈ 0.052 moles

Since the molecular formula of ethanol (C2H6O) contains 2 carbon atoms, there are twice as many moles of carbon atoms:
Number of moles of carbon atoms = 0.052 moles * 2 ≈ 0.104 moles

Therefore, there are approximately 0.104 moles of carbon atoms present in 2.4 g of ethanol.

2. To calculate the percentage of nitrogen by weight in ammonium sulfide, we need to determine the molar mass of nitrogen and ammonium sulfide. Then we can calculate the weight of nitrogen in ammonium sulfide and find the percentage.

First, let's calculate the molar mass of nitrogen (N):
Atomic mass of nitrogen (N) = 14.01 g/mol

Next, we will calculate the molar mass of ammonium sulfide (NH4)2S:
Molar mass = (2 * Atomic mass of Nitrogen) + (8 * Atomic mass of Hydrogen) + Atomic mass of Sulfur
= (2 * 14.01 g/mol) + (8 * 1.01 g/mol) + 32.07 g/mol
= 28.02 g/mol + 8.08 g/mol + 32.07 g/mol
= 68.17 g/mol

Now, we can calculate the weight of nitrogen in 3.696 mol of ammonium sulfide:
Weight of nitrogen = Number of moles of nitrogen * Molar mass of nitrogen
Weight of nitrogen = 2 * 3.696 mol * 14.01 g/mol
Weight of nitrogen = 103.42 g

To find the percentage of nitrogen by weight, we use the following formula:
% Nitrogen = (Weight of nitrogen / Weight of ammonium sulfide) * 100
% Nitrogen = (103.42 g / (3.696 mol * 68.17 g/mol)) * 100
% Nitrogen ≈ 55.82%

Therefore, the percentage of nitrogen by weight in 3.696 mol of ammonium sulfide is approximately 55.82%.

3. To calculate the number of atoms of mercury in liquid mercury, we need to determine the mass of mercury using its density. Then we can convert the mass to moles of mercury and finally calculate the number of atoms.

First, let's calculate the mass of mercury:
Mass = Volume * Density
Mass = 5.4 cm^3 * 13.55 g/cm^3
Mass ≈ 73.47 g

Next, we will calculate the number of moles of mercury using its molar mass:
Molar mass of mercury (Hg) = 200.59 g/mol (approx.)

Number of moles of mercury = Mass / Molar mass
Number of moles of mercury = 73.47 g / 200.59 g/mol ≈ 0.366 moles

Finally, we can use Avogadro's number to convert the moles of mercury to the number of atoms:
Number of atoms = Number of moles * Avogadro's number
Number of atoms = 0.366 moles * 6.022 × 10^23 atoms/mol ≈ 2.207 × 10^23 atoms

Therefore, there are approximately 2.207 x 10^23 atoms of mercury present in 5.4 cubic cm of liquid mercury.