Ask question

Ask question

All categories

3 years ago

14

A mixture of 1.20 mols He, 2.40 mols Ne, 4.80 mols Kr, and 0.60 mols Ar has a total pressure of 600.0 mm Hg. What is the partial

pressure of the Kr? Show all work to receive full credit.

1 answer:

Juliette [100K]3 years ago

8 0

Answer: The partial pressure of the Kr is 320 mm Hg.

Explanation:

According to Raoult's Law , the partial pressure of each component in the solution is equal to the total pressure multiplied by its mole fraction. It is mathematically expressed as

$p_A=x_A\times P_{total}$

where, $p_A$ = partial pressure of component A

$x_A$ = mole fraction of A

$P_{total}$ = total pressure

mole fraction of Krypton = $\frac{\text {Moles of Kr}}{\text {total moles}}=\frac{4.80}{1.20+2.40+4.80+0.60}=0.53$

$p_{Kr}=0.53\times 600mmHg=320mmHg$

Thus partial pressure of the Kr is 320 mm Hg

You might be interested in

1.) 3.4 moles Magnesium are mixed with 5.6 moles of Hydrochloric Acid. How many moles of Hydrogen gas are produced?

love history [14]

Answer:

1. 2.8 moles of H₂

2. 7.38 moles of CO₂

3. 5.3 moles of O₂

4. 7.4 moles of KNO₃

Explanation:

Here are the steps to doing this:

1. Write the chemical equation of each reaction.

2. Balance the equation.

3. Find out the ratio between reactant and product

4. Determine the actual yield of your reactants.

5. The amount of product produced is determined by how much product the limiting reactant produces.

Let's do this!

1. Given: 3.4 moles of Magnesium(Mg) and 5.6 moles of Hydrochloric acid (HCl)

<u>Equation:</u>

Mg + <u>2</u>HCl → MgCl₂ + H₂

<u>Reactant to Product ratio</u>

1 mole of Mg produces 1 mole of H₂ $\dfrac{1moleof Mg}{1mole of H_{2}}$

2 moles of HCl produces 1 mole of H₂ $\dfrac{2molesofHCl}{1mole of H_{2}}$

<u>Determine actual yield of reactants</u>

$3.4moles of Mg\times\dfrac{1moleofH_2}{1moleofMg}=3.4molesofH_{2}\\\\5.6moles ofHCl\times\dfrac{1moleofH_2}{2moleofHCl}=2.8molesofH_{2}$

Since 5.6 moles of HCl can only produce 2.8 moles of H₂, before it is used up, then this means that that is all the product this reaction can produce.

2. Given: 3.4 moles of C₃H₈ and 12.3 moles of oxygen gas (O₂)

<u>Equation:</u>

C₃H₈ + <u>5</u>O₂ → <u>3</u>CO₂ + <u>4</u>H₂O

<u>Reactant to Product ratio</u>

1 mole of C₃H₈ produces 3 moles of CO₂ $\dfrac{1moleofC_{3}H_{8}}{3molesofCO_{2}}$

5 moles of O₂ produces 3 moles of CO₂ $\dfrac{5molesofO_{2}}{3moleofCO_{2}}$

<u>Determine actual yield of reactants</u>

$3.4molesofC_{3}H_{8}\times\dfrac{3molesofCO_{2}}{1moleofC_{3}H_{8}}=10.2molesofH_{2}$

$12.3molesofO_{2}\times\dfrac{3molesofCO_{2}}{5molesofO_{2}}=7.38molesofCO_{2}$

The answer is then 7.38 moles of CO₂

**3. 5.3 moles of H₂O

This one is a little bit different. It is asking how much of a reactant is needed to produce the amount of product given. For this, just write a balanced equation for the reaction and get the ratio of reactant to product and solve for the actual yield. Since it is only asking for oxygen gas, you just need to do that one.

<u>Equation:</u>

CH₄ + <u>2</u>O₂ → CO₂ + <u>2</u>H₂O

<u>Reactant to Product ratio</u>

$\dfrac{2molesofO_{2}}{2molesofH_{2}O}=\[tex]7.88molesofKI\times\dfrac{1moleofKNO_{3}}{1moleofKI}=7.88molesofKNO_{3}$

<u>Actual yield:</u>

$5.3molesofH_{2}O\times\dfrac{1moleofO_{2}}{1moleofH_{2}O}=5.3molesofO_{2}$

The answer is 5.3 moles of O₂.

4. 3.7 moles of Lead (II) Nitrate (Pb(NO₃)₂) and 7.8 moles of Potassium Iodide (KI)

<u>Equation:</u>

Pb(NO₃)₂ + <u>2</u>KI → PbI₂ + <u>2</u>KNO₃

<u>Reactant to Product ratio</u>

$\dfrac{2molesofKI}{2molesofKNO_{3}}=\dfrac{1moleofKI}{1moleofKNO_{3}}$

$\dfrac{1molesofPb(NO_{3})_{2}}{2molesofKNO_{3}}$

<u>Actual yield:</u>

$3.7molesofPb(NO_{3})_{2}\times\dfrac{2molesofKNO_{3}}{1moleofPb(NO_{3})_{2}}=7.4molesofKNO_{3}$

The answer is 7.4 moles of KNO₃.

3 0

3 years ago

Can someone answer that first this pls im lost

Alchen [17]

-Just look up “H2O lewis structure
-1.5
-Don’t know the VSEPR
-Polar Covalent
-Again, don’t know VSEPR
-Just look up H2O molecule

4 0

3 years ago

A gas has a volume of 5.00 L at 0°C. What final temperature, in degrees Celsius, is needed to change the volume of the gas to ea

Diano4ka-milaya [45]

Answer:

A = -213.09°C

B = 15014.85 °C

C = -268.37°C

Explanation:

Given data:

Initial volume of gas = 5.00 L

Initial temperature = 0°C (273 K)

Final volume = 1100 mL, 280 L, 87.5 mL

Final temperature = ?

Solution:

Formula:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Conversion of mL into L.

Final volume = 1100 mL/1000 = 1.1 L

Final volume = 87.5 mL/1000 = 0.0875 L

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 1.1 L × 273 K / 5.00 L

T₂ = 300.3 L.K / 5.00 K

T₂ = 60.06 K

60.06 K - 273 = -213.09°C

2)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 280 L × 273 K / 5.00 L

T₂ = 76440 L.K / 5.00 K

T₂ = 15288 K

15288 K - 273 = 15014.85 °C

3)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

4.78 K - 273 = -268.37°C

4 0

3 years ago

Marianna [84]

Answer:

B

Explanation:

Trust me I've had problems like these

6 0

3 years ago

Read 2 more answers

The following reaction shows calcium chloride reacting with silver nitrate.

adell [148]

Grams of Ca(NO₃)₂ produced : 0.985 g

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

CaCl₂ + 2AgNO₃ → 2AgCl + Ca(NO₃)₂

MW AgNO₃ : 107.9+14+3.16=169.9

mol AgNO₃ :

$\tt \dfrac{2}{169.9}=0.012$

mol ratio Ca(NO₃)₂ : AgNO₃ = 1 : 2, so mol Ca(NO₃)₂ :

$\tt \dfrac{1}{2}\times 0.012=0.006$

MW Ca(NO₃)₂ : 40.1+2.14+6.16=164.1 g/mol

mass Ca(NO₃)₂ :

$\tt 0.006\times 164.1=0.985$

7 0

3 years ago