1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Zina [86]
2 years ago
7

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

the following is false?
Select one:

A. The final concentration of Pb2+ ions is 0.174 M.

B. You form 111 g of lead(II) iodide.

C. The final concentration of K+ is 0.821 M.

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

E. All are true.
Chemistry
1 answer:
SIZIF [17.4K]2 years ago
4 0

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

<u>Moles of potassium iodide = 0.48 moles </u>

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

<u>Moles of lead(II) nitrate  = 0.342 moles </u>

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

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

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

<u>So, Concentration = 0.102/0.585 M = 1.174 M</u>

<u>Statement A is correct.</u>

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

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

<u>Statement B is correct.</u>

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

<u>So, Concentration = 0.48/0.585 M = 0.821 M</u>

<u>Statement C is correct.</u>

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

<u>So, Concentration = 0.684/0.585 M = 1.169 M</u>

<u>Statement D is incorrect.</u>

You might be interested in
The thermite reaction reacts iron (III) oxide, Fe2O3, with aluminium powder,Al, the form aluminium oxide, Al2O3 and iron, Fe.
lara [203]

Answer:

<u><em>This answer assumes that the strated "16.0g of iron" was meant to be 16.0 grams of iron(III) oxide.</em></u>

Explanation:

To start, the thermite equation must be balanced.

I find:

1Fe2O3 + 2Al = 1Al2O3 + 2Fe

This tells us we need 2 moles of Al for every 1 mole of Fe2O3.  

Now calculate the moles of each reactant:

Moles Fe2O3:  16.0 g/159.7 g/mole = <u>0.100 moles Fe2O3</u>

Moles Al:  8.1 /26.98 g/mole = <u>0.300 moles Al</u>

The balanced equation says that in order to react all of the Fe2O3 we'd need twice that amount (in moles) of the Al.  (0.100 moles Fe2O3)*(2) = 0.200 moles Al.

<u>Which of the two reactants is the limiting reagent?</u>

We have more than enough moles of Al to react with 0.10 moles of Fe2O3.  (We have 0.300 moles Al and all we need is 0.200 moles to react with the 0.10 moles of Fe2O3.  <em>Fe2O3 is the limiting reagent.</em>

<u><em>Calculate the maximum mass of iron of iron that could be formed using these quantities of reactants.</em></u>

The balanced equation tells us that we will obtain 2 moles of Fe for every 1 mole of Fe2O3 consumed.  Since Fe2O3 is the limiting reagent, we will assume that it completely reacts.  That means 0.1 moles of Fe2O3 is reacted.  Since we expect twice that many moles of Fe, we should obtain 0.200 moles of Fe.  At 55.85 g/mole, we should obtain:

(0.200 moles Fe)*(55.85 g Fe/mole Fe) = 11.2 grams Fe

<em></em>

<em></em>

5 0
1 year ago
1. Why does light have different effects on various materials?
Debora [2.8K]
Light basically evaporates pigments such as colored paper. Exposed pigment fading is kind of like getting colored paper wet. It will fade and bleed.
4 0
2 years ago
Read 2 more answers
Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov
love history [14]

Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field  = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3

Mass of the Astroturf = m

Density of the Astroturf = d = 187 oz/ft^3

d=\frac{m}{V}

m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz

1 oz = 0.0283495 kg

m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg

Weight of the Astroturf = W

W = mg

=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N

The weight of the Astroturf is 179,684.31 Newtons.

8 0
3 years ago
A tube of air contains a feather and a coin. both objects are dropped at the same time. the coin falls faster than the feather.
Mazyrski [523]
Because gravity treats all mass equally. It doesn't treat heavier objects with greater force. All objects experience the gravitational pull equally. Making the two different weights reach the ground together simultaneously despite their weight differences.
6 0
3 years ago
What is the atomic number of beryllium
Alexandra [31]

Answer:

4 the symbol is Be

Explanation:

hope this helped

8 0
3 years ago
Read 2 more answers
Other questions:
  • For the reaction: CH3NH2(aq) + H2O(aq) ⇌ CH3NH3 +(aq) + OH- Determine the change in the pH (ΔpH) for the addition of 6.7 M CH3NH
    14·1 answer
  • Marcie is hiking in the mountains and she discovers an interesting shiny rock.As she picks it up she wonders whether it could be
    5·1 answer
  • Give the formula for the compound formed when sulfur trioxide reacts with water.
    6·1 answer
  • A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net
    13·1 answer
  • What is the pressure of the air in a car tire (in psi) if it is 1.73 atm?
    8·1 answer
  • PLEASE HELP WORTH 50 POINTS NO LINKS I WILL MARK CORRECT ANSWER BRAINLIEST!!! complete this punnet square, enter your answer in
    10·1 answer
  • PLS HELP!! WILL GIVE BRAINLY!!
    11·1 answer
  • Could Rutherford make any conclusions about electrons based on the result of experiment?
    10·1 answer
  • 18) What is the following number representing: 10m/s West *
    10·1 answer
  • In what way is the Moon like the Earth?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!