From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of fluorine = 18.99 grams
molecular mass of chlorine = 35.5 grams
Therefore:
one mole of CF2Cl2 = 12 + 2(18.99) + 2(35.5) = 120.98 grams
Therefore, we can use cross multiplication to find the number of moles in 79.34 grams as follows:
mass = (79.34 x 1) / 120.98 = 0.6558 moles

Now, one mole contains 6.022 x 10^23 molecules, therefore:
number of molecules in 0.65548 moles = 0.6558 x 6.022 x 10^23
= 3.949 x 10^23 molecules

7 0

3 years ago

Suppose a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O). The total pressure of

julia-pushkina [17]

Considering the definition of mole fraction, the mole fraction of O₂ in the mixture is 0.434.

<h3>Definition of mole fraction</h3>

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

In other words, the mole fraction expresses the concentration of solute in a solution as the ratio of moles of substance to total moles of solution:

$mole fraction=\frac{moles of substance}{moles of solution}$

<h3>Mole fraction of O₂ in this mixture</h3>

In this case, you know a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O).

So, the total moles of the solution can be calculated as:

Total moles = moles of oxygen (O₂) + moles of nitrous oxide (N₂O)

Then:

Total moles= 4.60 moles + 6 moles

Total moles= 10.60 moles

Finally, the more fraction of O₂ can be calculated as follow:

$Mole fraction of O_{2} =\frac{moles of O_{2}}{total moles}$

$Mole fraction of O_{2} =\frac{4.60 moles}{10.6o moles}$

Solving:

<u><em>Mole fraction O₂ = 0.434</em></u>

Finally, the mole fraction of O₂ in the mixture is 0.434.

Learn more about mole fraction:

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3 0

2 years ago

How do the given animal reproduce

Aliun [14]

Answer:

Which animals????????

3 0

3 years ago

Read 2 more answers

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that

iragen [17]

Answer:

108.43 grams KNO₃

Explanation:

To solve this problem we use the formula:

ΔT = Kf * b * i

Where

ΔT is the temperature difference (14.5 K)

Kf is the cryoscopic constant (1.86 K·m⁻¹)

b is the molality of the solution (moles KNO₃ per kg of water)

and<em> i</em> is the van't Hoff factor (2 for KNO₃)

We <u>solve for b</u>:

14.5 K = 1.86 K·m⁻¹ * b * 2

b = 3.90 m

Using the given volume of water and its density (aprx. 1 g/mL) we <u>calculate the necessary moles of KNO₃</u>:

275 mL water ≅ 275 g water

275 g /1000 = 0.275 kg

moles KNO₃ = molality * kg water = 3.90 * 0.275

moles KNO₃ = 1.0725 moles KNO₃

Finally we <u>convert KNO₃ moles to grams</u>, using its molecular weight:

1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃

5 0

2 years ago