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professor190 [17]
3 years ago
15

B. What do you notice about the number of atoms in one mole?

Chemistry
1 answer:
Doss [256]3 years ago
5 0

Answer:

one mole of atom of any element contains6.022×1033 atoms regardless of the type of elements the mass of one mole of an element depend on what that element is and is equal to atom mass of that element in gram

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What is the total pressure in a container with 256 mm Hg of Oz, 198 mm Hg of He,
irina1246 [14]

Answer:

P(total pressure) = 504 mmHg = 504mm/760mm/atm = 0.663 atm

Explanation:

Apply Dalton's Law of Partial Pressures.

P(total) = ∑Partial Pressures = ∑(256mm + 198mm + 48mm) = 504 mmHg

P(total pressure) = 504 mmHg = 504mm/760mm/atm = 0.663 atm

7 0
3 years ago
Be sure to answer all parts. Calculate the molality, molarity, and mole fraction of FeCl3 in a 24.0 mass % aqueous solution (d =
Anna71 [15]

Answer:

m= 1.84 m

M= 1.79 M

mole fraction (X) =

Xsolute= 0.032

Xsolvent = 0.967

Explanation:

1. Find the grams of FeCl3 in the solution: when we have a mass % we assume that there is 100 g of solution so 24% means 24 g of FeCl3 in the solution. The rest 76 g are water.

2. For molality we have the formula m= moles of solute / Kg solvent

so first we pass the grams of FeCl3 to moles of FeCl3:

24 g of FeCl3x(1 mol FeCl3/162.2 g FeCl3) = 0.14 moles FeCl3

If we had 76 g of water we convert it to Kg:

76 g water x(1 Kg of water/1000 g of water) = 0.076 Kg of water

now we divide m = 0.14 moles FeCl3/0.076 Kg of water

m= 1.84 m

3. For molarity we have the formula M= moles of solute /L of solution

the moles we already have 0.14 moles FeCl3

the (L) of solution we need to use the density of the solution to find the volume value. For this purpose we have: 100 g of solution and the density d= 1.280 g/mL

The density formula is d = (m) mass/(V) volume if we clear the unknown value that is the volume we have that (V) volume = m/d

so V = 100 g / 1.280 g/mL = 78.12 mL = 0.078 L

We replace the values in the M formula

M= 0.14 moles of FeCl3/0.078 L

M= 1.79 M

3. Finally the mole fraction (x)  has the formula

X(solute) = moles of solute /moles of solution

X(solvent) moles of solvent /moles of solution

X(solute) + X(solvent) = 1

we need to find the moles of the solvent and we add the moles of the solute like this we have the moles of the solution:

76 g of water x(1 mol of water /18 g of water) = 4.2 moles of water

moles of solution = 0.14 moles of FeCl3 + 4.2 moles of water = 4.34 moles of solution

X(solute) = 0.14 moles of FeCl3/4.34 moles of solution = 0.032

1 - X(solute) = 1 - 0.032 = 0.967

6 0
3 years ago
What is the mole fraction of each component if 3.9 g of benzene (C6H6) is dissolved in 4.6 g of toluene (C7H8)
Savatey [412]

Answer:

Step 1 of 6

(a)

The mass of benzene is  , so calculate the moles of benzene as follows:



The mass of toluene is, so calculate the moles of toluene as follows:



Now, calculate the mole fraction as follows:





Therefore, the mole fraction of benzene and toluene is  and  respectively.

Step 2 of 6

(b)

The formula to calculate the partial pressure is as follows:



Here,  is the partial pressure of benzene,  is the vapour pressure of pure benzene and  is the mole fraction of benzene.

Vapour pressure of pure benzene at  is.

Substitute the values in the equation as follows:



Therefore, the partial pressure is  .

Step 3 of 6

(c)

Vapor pressure of the solution at 1 atm is  .

When the total pressure of the vapour pressure of the mixture is  at a temperature, then, the solution boils. It corresponds to the boiling point of the solution.

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is less than the atmospheric pressure, the solution will not boil at  .

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is greater than the atmospheric pressure, the solution will boil at  .

Therefore, the boiling point of the solution is  .

Step 4 of 6

(d)

Mole fraction of benzene at  is calculated as follows:



Mole fraction of toluene at  is calculated as follows:



Therefore, the mole fractions of benzene and toluene are  and  respectively.

Step 5 of 6

(e)

Vapor pressure of benzene at  is  .

Partial pressure of benzene is calculated as follows:



Vapor pressure of toluene at  is  .

Partial pressure of toluene is calculated as follows:



Step 6 of 6

Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Explanation:

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