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2 years ago

Determine the number of atoms in 1.85 ml of mercury. (the density of mercury is 13.5 g/ml.)

1 answer:

7 0

First step is to get the mass of the mercury:
Pressure = mass/volume
mass = pressure x volume = 13.5 x 1.85 = 24.975 gm

Second step is to calculate the number of moles in 24.975 gm:
From the periodic table, the molar mass of mercury is 200.59
mass = number of moles x molar mass
number of moles = 24.975 / 200.59 = 0.1245 mole

Last step is to get the number of atoms:
Each mole contains 6.02 x 10^23 atoms
number of atoms = 0.1245 x 6.02 x 10^23 = 7.4949 x 10^22 atoms

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A common factor is low pressure system.

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If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?

Arada [10]

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

Answer:

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2 years ago

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Which solution contains the largest number of moles of chloride ions?

elixir [45]

Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution

Part (a): 30.00 ml of 0.100m Cacl2
number of moles of CaCl2 = 0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine

Part (b): 10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine

Part (c): 4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine

Part (d): 7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine

Based on the above calculations, the correct answer is (d)

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2 years ago

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Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

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1 year ago

A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a

Nata [24]

Answer:

There are present 5,5668 moles of water per mole of CuSO₄.

Explanation:

The mass of CuSO₄ anhydrous is:

23,403g - 22,652g = 0,751g.

mass of crucible+lid+CuSO₄ - mass of crucible+lid

As molar mass of CuSO₄ is 159,609g/mol. The moles are:

0,751g × $\frac{1mol}{159,609g}$ = 4,7052x10⁻³ moles CuSO₄

Now, the mass of water present in the initial sample is:

23,875g - 0,751g - 22,652g = 0,472g.

mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid

As molar mass of H₂O is 18,02g/mol. The moles are:

0,472g × $\frac{1mol}{18,02g}$ = 2,6193x10⁻² moles H₂O

The ratio of moles H₂O:CuSO₄ is:

2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668

That means that you have 5,5668 moles of water per mole of CuSO₄.

I hope it helps!

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2 years ago