How does amount of greenhouse gases affect the temperature of the Earth's atmosphere?

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

2 years ago

What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances

belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope mass amu Relative abundance

1 77.9 14.4

2 81.9 14.3

3 85.9 71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9

% abundance of isotope 1 = 14.4% = $\frac{14.4}{100}=0.144$

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = $\frac{14.3}{100}=0.143$

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = $\frac{71.3}{100}=0.713$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]$

$A=84.2amu$

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0

3 years ago

You run a 5k for charity. How many feet do you run?

slamgirl [31]

You run approximately 16404 feet.

3 0

2 years ago

A balloon is filled with 2.5 L of helium at 1 atm of pressure. If the pressure

ch4aika [34]

Answer:

5 L

Explanation:

Given data

Initial pressure (P₁): 1 atm

Initial volume (V₁): 2.5 L

Final pressure (P₂): 0.50 atm

Final volume (V₂): ?

For a gas, there is an inverse relationship between the pressure and the volume. Mathematically, for an ideal gas that undergoes an isothermic change, this is expressed through Boyle's law.

$P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{1atm \times 2.5L}{0.50atm}= 5 L$

8 0

3 years ago

Question 9

Evgesh-ka [11]

Answer:

0.382 atm

Explanation:

In order to find the pressure, you need to know the moles of carbon dioxide (CO₂) gas. This can be found by multiplying the mass (g) by the molar mass (g/mol) of CO₂. It is important to arrange the conversion in a way that allows for the cancellation of units.

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

15 grams CO₂ 1 mole
---------------------- x ------------------------ = 0.341 moles CO₂
44.007 grams

To find the pressure, you need to use the Ideal Gas Law equation.

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

After you convert Celsius to Kelvin, you can plug the given and calculated values into the equation and simplify to find the pressure.

P = ? atm R = 0.08206 atm*L/mol*K

V = 20 L T = 0 °C + 273.15 = 273.15 K

n = 0.341 moles

PV = nRT

P(20 L) = (0.341 moles)(0.08206 atm*L/mol*K)(273.15 K)

P(20 L) = 7.64016

P = 0.382 atm

7 0

2 years ago