How many moles of precipitate will be formed when 100.0 mL of 0.200 M NaBr is reacted with excess Pb(NO₃)₂ in the following chemical reaction?

2 NaBr (aq) + Pb(NO₃)₂ (aq) → PbBr₂ (s) + 2 NaNO₃ (aq)

Answers

Answer 1

Answer : The number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.

Explanation : Given,

Concentration of NaBr = 0.200 M

Volume of solution = 100.0 mL = 0.1 L      (1 L = 1000 mL)

First we have to calculate the moles of NaBr.

[tex]\text{Moles of NaBr}=\text{Concentration of NaBr}\times \text{Volume of solution in L}[/tex]

[tex]\text{Moles of NaBr}=0.200M\times 0.1L=0.02mol[/tex]

Now we have to calculate the moles of precipitate, [tex]PbBr_2[/tex] formed.

The balanced chemical reaction is:

[tex]2NaBr(aq)+Pb(NO_3)_2(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)[/tex]

From the balanced chemical reaction we conclude that:

As, 2 moles of NaBr react to give 1 mole of [tex]PbBr_2[/tex]

So, 0.02 moles of NaBr react to give [tex]\frac{0.02}{2}=0.01[/tex] mole of [tex]PbBr_2[/tex]

Therefore, the number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.

Answer 2

The number of mole of the precipitate (i.e PbBr₂) formed when 100 mL of 0.2 M NaBr react with excess Pb(NO₃)₂ is 0.01 mole

We'll begin by calculating the number of mole of NaBr in 100 mL of 0.2 M NaBr solution. This can be obtained as follow:

Volume = 100 mL = 100 / 1000 = 0.1 L

Molarity of NaBr = 0.2 M

Mole of NaBr =?

Mole = Molarity x Volume

Mole of NaBr = 0.2 × 0.1

Mole of NaBr = 0.02 mole

Finally, we shall determine the number of mole of the precipitate (i.e PbBr₂) produced from the reaction. This can be obtained as follow:

2NaBr(aq) + Pb(NO₃)₂(aq) → PbBr₂(s) + 2NaNO₃ (aq)

From the balanced equation above,

2 moles of NaBr reacted to produce 1 mole of PbBr₂.

Therefore,

0.02 mole of NaBr will react to produce = [tex]\frac{0.02}{2} \\\\[/tex] = 0.01 mole of PbBr₂.

Thus, the number of mole of the precipitate (i.e PbBr₂) produced from the reaction is 0.01 mole

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Which of the following is a good definition of matter?
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Answer:

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Answer :
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Answer:

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Answers

Answer:

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Collect like terms

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Answer:

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Question 11
4 pts
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Answers

Answer:

3.65 mol O₂

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Answers

Answer:

you can see the empirical formula at the pic

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What is empirical formula?

Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.

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(b) 2.677 g Ba : 3.115 g Br

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Answer:

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The test variable (independent variable) controls the outcome variable (dependent variable)

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its right on study island

SOMEONE PLEASE HELPPP​

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A is the correct awnser Beacuse it ether right kne

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Answers

Answer:

[tex]Total = 50.6\ moles[/tex]

Explanation:

Given

[tex]Propane = C_3H_8[/tex]

Represent Carbon with C and Hydrogen with H

[tex]C = 13.8[/tex]

Required

Determine the total moles

First, we need to represent propane as a ratio

[tex]C_3H_8[/tex] implies

[tex]C:H = 3:8[/tex]

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Substitute 13.8 for C

[tex]13.8 : H = 3 : 8[/tex]

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[tex]\frac{13.8}{H} = \frac{3}{8}[/tex]

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A sample of an unknown gas weighs 0.419 grams and produced 5.00L of gas at 1.00atm (unknown gas only) and 298.15K, what is the molar mass (g/mole) of this unknown gas

Answers

Answer:

molar mass of unknown gas = 1.987 g/mol

Explanation:

First, the number of moles of the unknown gas is found

Using the ideal gas equation: PV = nRT

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

n = PV/RT

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

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molar mass = 0.419 g/ 0.2109 mols

molar mass of unknown gas = 1.987 g/mol

The molar mass of unknown gas by using ideal gas equation = 1.987 g/mol.

Ideal gas equation

This equation gives the relation between pressure, volume, temperature as given below:

[tex]PV = nRT[/tex]

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

Substitute the above values in the above equation as follows:

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

Formula for molar mass

[tex]Molar mass = mass/ number of moles[/tex]

Calculate molar mass by using the above equation,

molar mass = 0.419 g/ 0.2109 moles

The molar mass of unknown gas = 1.987 g/mol

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brainly.com/question/4147359

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heeeeeeeeeelp please please please ​

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Answer:

Explanation:

In my opinion the answer should be SO2

Answer:

a should be answer i think.

Scientists are experimenting with pure samples of isotope X which is radioactive. The sample has a mass of 20. Grams. The half-life was measured to be 232 seconds. There is a second sample that weighs 80 grams. What is the half-life of the second sample

Answers

Answer:

Explanation:

Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .

Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .

calculate the mass of N2 gas which has a volume 0.227 at STP​

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Hi I just wanted you know I have to work at the gym I will call when you I have a good night love I will

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Answers

During sublimation

This has been posted on here before so you could’ve searched it lol.

Best of luck :))

Answer:

during sublimation

Explanation:

just took the test

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Answers

Answer:

The correct answer is 0.67 g H₂

Explanation:

Isopropyl alcohol (C₃H₇OH) can decompose to give acetone (C₂H₆OH) and hydrogen gas (H₂) according to the following chemical equation:

C₃H₇OH (g) ⇒ C₂H₆CO(g) + H₂(g)

We can calculate the initial mass of isopropyl alcohol from the density and volume data:

density = m/V = 0.785 g/mL

⇒ m = density x V = 0.785 g/mL x 25.6 mL = 20.096 g C₃H₇OH

According to the chemical equation 1 mol of C₃H₇OH gives 1 mol H₂. The molar mass of C₃H₇OH is:

molar mass C₃H₇OH = (12 g/mol x 3) + (1 g/mol x 7) + 16 g/mol + 1 g/mol = 60 g/mol

molar mass H₂ = 1 g/mol x 2 = 2 g/mol

So, we obtain: 2 g H₂ from 60 g C₃H₇OH. We multiply this stoichiometric ratio (2 g H₂/60 g C₃H₇OH) by the initial mass of C₃H₇OH to obtain the mass of H₂ is formed:

20.096 g C₃H₇OH x (2 g H₂/60 g C₃H₇OH) = 0.6698 g ≅ 0.67 g H₂

True or False: Particles that are moving faster have a higher temperature

Answers

Answer:

true

Explanation:

I'm not sure why cause I dont know how to explain but it's TRUE

Answer:

True  

Explanation:

The particles moving faster in a substance the hotter it gets.

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