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Calculate the number of moles of C2H6 in 6.29×1023 molecules of C2H6.

Kipish [7]

1.05moles

Explanation:

Given parameters:

Number of molecules of C₂H₆ = 6.29 x 10²³molecules

Unknown:

Number of moles = ?

Solution:

The mole is the amount of substances that contains Avogadro's number of particles i.e 6.02 x 10²³

To find the number of moles:

number of moles = $\frac{number of particles}{Avogadro's number}$

number of moles = $\frac{6.29 x 10^{23} }{6.02 x 10^{23} }$

number of moles = 1.05moles

Learn more:

moles brainly.com/question/1841136

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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

The Society of Automotive Engineers has established an accepted numerical scale to measure the viscosity of motor oil. For examp

natta225 [31]

Answer:

Most viscous to least viscous: $(c)> (b)> (a)$

Explanation:

For hydrocarbons, viscosity increases with increasing molar mass. Because increasing molar mass signifies increase in number of electrons in molecules.

We know that in non-polar hydrocarbons, only van der waal intermolecular force exists. Van der waal force is proportional to number of electrons in a molecule.

Therefore with increasing molar mass, van der waal force increases. hence molecules gets more tightly bind with each other resulting increase in viscosity.

Here molar mass order : $(c)> (b)> (a)$

Therefore viscosity order : $(c)> (b)> (a)$

5 0

3 years ago

How many moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas?

tino4ka555 [31]

Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CO₂ + 4 H₄ → CH₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CO₂: 1 mole
H₄: 4 moles
CH₄: 1 mole
H₂O: 2 moles

<h3>Moles of CH₄ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

$moles of CH_{4} =\frac{85.1 moles of CO_{2}x4 moles of CH_{4} }{1 moles of CO_{2}}$