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1 year ago

determine the molar mass (in g/mol) of a gas that travels with an effusion time that is 1.94 times longer than that of hcl(g).

1 answer:

8 0

The molar mass of a gas that travels with an effusion time that is 1.94 times longer than that of HCl (g) is 137 g/mol

<h3>What is the relationship between the molar mass of a gas and the effusion time of a gas?</h3>

The relationship between the molar mass of a gas and the effusion time of a gas is given by the equation below:

T₁/T₂ = √M₁/√M₂

where;

T₁ is the time of effusion of HCl

T₂ is the time of effusion of the gas

M₂ is the molar mass of the gas

M₁ is the molar mass of HCl = 36.5 g/mol

T₂ = 1.94T₁

1 / 1.94 = √36.5 /√M₂

M₂ = 137 g.mol

Learn more about the effusion time at: brainly.com/question/22359712

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Q = 0.75 g x 0.897 J/g•°C x 22°C

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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

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2 years ago

Place the following in order of increasing acid strength. HBrO2 HBrO3 HBrO HBrO4 A) HBrO2

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Answer:

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Mrac [35]

E(Bonds broken) = 1371 kJ/mol reaction
E(Bonds formed) = 1852 kJ/mol reaction

ΔH = -481 kJ/mol.
The reaction is exothermic.

<h3>Explanation</h3>

2 H-H + O=O → 2 H-O-H

There are two moles of H-H bonds and one mole of O=O bonds in one mole of reactants. All of them will break in the reaction. That will absorb

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ΔH(Breaking bonds) = +1371 kJ/mol

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E(Bonds formed) = 2 × 2 × 463 = 1852 kJ/mol reaction.
ΔH(Forming bonds) = - 1852 kJ/mol

Heat of the reaction:

$\Delta H_{\text{rxn}} = \Delta H(\text{Breaking bonds}) + \Delta H(\text{Forming bonds})\\\phantom{ \Delta H_{\text{rxn}}} = +1371 + (-1852) \\\phantom{ \Delta H_{\text{rxn}}} = -481 \; \text{kJ} / \text{mol}$

$\Delta H_{\text{rxn}}$ is negative. As a result, the reaction is exothermic.

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