7) Facilitated diffusion is

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

3 years ago

if you are told to get 100 mL of stock solution to use to prepare smaller size sample for an experiment, which piece of glasswar

fgiga [73]

Answer:

A beaker

Step-by-step explanation:

Specifically, I would use a 250 mL graduated beaker.

A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.

You don't need precisely 100 mL solution.

If the beaker is graduated, you can easily measure 100 mL of the stock solution.

Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).

6 0

2 years ago

At the heart of the stars, nuclear fusion merges _____ together in the nucleus of atoms to create new elements?

Goshia [24]

Answer:

lighter atoms especially hydrogen

Explanation:

At the heart of the stars, nuclear fusion merges lighter atoms especially hydrogen together in the nucleus of atoms to create new elements.

During nuclear fusion small atomic nuclei combines to form larger ones with the release of a large amount of energy.

The energy released provides the needed temperature for another set of hydrogen atoms to fuse. This process is in turn yields another set of helium atom which releases a lot of energy.

A chain reaction progresses which leads to the formation of new elements.

3 0

3 years ago

How many milliliters of 0.0510 m edta are required to react with 50.0 ml of 0.0200 m cu2 ?

krok68 [10]

Cu ions plus EDTA2- ->cu(EDTA)2- plus 2H-
number of moles of CU ions used which is equal to molarity multiplied by volume in litres
that is 50xo.o2 divided by 1000 that is 0.001moles
Since ratio is 1:1 the moles of EDTA is also0.001moles
volume of EDTA is 0.001 divided by 0.0510m which is 0.0196 litres in ml is 0.0196x1000 which is 19.61ml

4 0

3 years ago

The chemical formula for magnesium oxide is 2MgO. The number 2 represents the number of ___.

patriot [66]

Answer:

Magnesium oxide molecules

Explanation:

3 0

3 years ago