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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Answers

Answer 1

Answer:

Answer:

Mass PbI2 = 18.19 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams

Answer 2

Answer:

Answer:

$m_(PbI_2)=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_(Pb(NO_3)_2)=(0.14gPb(NO_3)_2)/(1g\ sln)*(1molPb(NO_3)_2)/(331.2gPb(NO_3)_2) *(1.134g\ sln)/(1mL\ sln) *96.7mL\ sln\n\nn_(Pb(NO_3)_2)=0.04635molPb(NO_3)_2\n\nn_(KI)=(0.12gKI)/(1g\ sln)*(1molKI)/(166.0gKI) *(1.093g\ sln)/(1mL\ sln) *99.8mL\ sln\n\nn_(KI)=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*(2molKI)/(1molPb(NO_3)_2) =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_(PbI_2)=0.07885molKI*(1molPbI_2)/(2molKI) *(461.01gPbI_2)/(1molPbI_2) \n\nm_(PbI_2)=18.2gPbI_2$

Best regards.

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A student is heating a chemical in a beaker with a Bunsen burner.In a paragraph of at least 150 words, identify the safety equipment that should be used and the purpose of it for the given scenario.

Answers

When a student is warming a chemical in a container using a special burner, it is very important to focus on safety by using the right safety tools.

What is the safety equipment

First, the student needs to wear the right safety clothes like a lab coat, gloves, and goggles to protect themselves from getting splashed or hurt by chemicals. A lab coat stops chemicals from touching the skin, gloves keep the hands safe, and safety goggles protect the eyes from chemicals

and hot things.

Furthermore, using a fume hood is necessary to make sure there is enough fresh air circulating and to remove any dangerous fumes or gases that might be released while heating things up.

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Answer:The student should be wearing a lab coat or maybe an apron to prevent chemicals from spilling or exploding onto their clothes, I do recommend a lab coat better though because it can protect your skin better. Next, make sure while messing with chemicals you are always wearing goggles, if you are not wearing them there is a chance that after touching chemicals you could touch your eyes. And that brings me to washing your hands straight away after messing with chemicals. You could also wear gloves and just take them off when you're done but if you don't have clean hands afterward you could always put the chemicals all over your skin. But in case you do touch your eyes there is always an emergency eyewash station somewhere in the lab room. And if you are to get Chemicals on your skin, in your hair, on your clothes, or to be on fire, there shall be a shower somewhere to get rid of that. But if you read the instructions or listen closely to the teacher you shall have no problem.

Explanation:

I kinda got off topic

For each of the following redox reactions in the, identify the oxidizing agent and the reducing agent.2Na(aq)+2H2O(l)→2NaOH(aq)+H2(g)
C(s)+O2(g)→CO2(g)
2MnO−4(aq)+5SO2(g)+2H2O(l)→2Mn2+(aq)+5SO2−4(aq)+4H+(aq)

Answers

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Gasline is solid by the liter in many countries. How many liters are required to fill a 12.0-gal gas tank?

Answers

Answer : The volume required to fill the gas tank is, 45.42 liters

Explanation :

Conversion used for gallon to liters are:

$1\text{ gallon}=3.785\text{ liter}$

As we are given the volume of gas tank in gallon is, 12.0 gal

Now we have to determine the volume of gas tank in liters.

As, 1 gallon = 3.785 liter

So, 12.0 gallon = $\frac{12.0\text{ gallon}}{1\text{ gallon}}* 3.785\text{ liter}=45.42\text{ liter}$

Therefore, the volume required to fill the gas tank is, 45.42 liters

What is the purpose of a catalyst?O A. To change the potential energy of the reactants
O B. To lower the activation energy of a reaction
O C. To increase the kinetic energy of the reactants
O D. To shift the equilibrium position of a reaction

Answers

B. To lower the activation energy of a reaction

Answer:

To lower the activation energy of a reaction

Explanation:

i just took the test and got it right ...... i hope this helps :)

Consider the following unbalanced equation for the combustion of hexane: αC6H14(g)+βO2(g)→γCO2(g)+δH2O(g) Part A Balance the equation. Give your answer as an ordered set of numbers α, β, γ, ... Use the least possible integers for the coefficients. α α , β, γ, δ = nothing Request Answer Part B Determine how many moles of O2 are required to react completely with 5.6 moles C6H14. Express your answer using two significant figures. n n = nothing mol Request Answer Provide Feedback

Answers

Answer:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

α =2

β = 19

γ = 12

δ = 14

53.2moles of O₂

Explanation:

Proper equation of the reaction:

αC₆H₁₄ + βO₂ → γCO₂ + δH₂O

This is a combustion reaction for a hydrocarbon. For the combustion of a hydrocarbon, the combustion equation is given below:

CₓHₙ + (x + $(n)/(4)$ )O₂ → xCO₂ + $(n)/(2)$ H₂O

From the given combustion equation, x = 6 and n = 14

Therefore:

β = x + $(n)/(4)$ = 6 + $(14)/(4)$ = 6 + 3.5 = 9 $(1)/(2)$

γ = 6

δ = $(n)/(2)$ = $(14)/(2)$ = 7

The complete reaction equation is therefore given as:

C₆H₁₄ + 9 $(1)/(2)$ O₂ → 6CO₂ + 7H₂O

To express as whole number integers, we multiply the coefficients through by 2:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

Problem 2

From the reaction:

2 moles of hexane are required to completely react with 19 moles of O₂

∴ 5.6 moles of hexane would react with k moles of O₂

This gives: 5.6 x 19 = 2k

k = $(5.6 x 19)/(2)$

k = 53.2moles of O₂

Write a net ionic equation for the reaction that occurs when excess hydrochloric acid (aq) and potassium sulfite (aq) are combined. Note: Sulfites follow the same solubility trends as sulfates.

Answers

Answer: The net ionic equation for the given reaction is $2H^+(aq.)+SO_3^(2-)(aq.)\rightarrow H_2O(l)+SO_2(g)$

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are the ions which do not get involved in a chemical equation. It is also defined as the ions that are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:

$2HCl(aq.)+K_2SO_3(aq.)\rightarrow 2KCl(aq.)+SO_2(g)+H_2O(l)$

Ionic form of the above equation follows:

$2H^+(aq.)+2Cl^-(aq.)+2K^+(aq.)+SO_3^(2-)(aq.)\rightarrow 2K^+(aq.)+2Cl^-(aq.)+SO_2(g)+H_2O(l)$

As, potassium and chloride ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$2H^+(aq.)+SO_3^(2-)(aq.)\rightarrow SO_2(g)+H_2O(l)$

Hence, the net ionic equation for the given reaction is written above.

Final answer:

The net ionic equation for the reaction between hydrochloric acid and potassium sulfite is H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq), following the solubility trends of sulfates and sulfites under standard conditions.

Explanation:

The reaction between excess hydrochloric acid (HCl) and potassium sulfite (K2SO3) is a typical acid-base neutralization reaction. In the initial step, potassium sulfite dissociates into its ions in the aqueous solution:

K2SO3 (aq) → 2K+ (aq) + SO3^2- (aq)

Hydrochloric acid, being a strong acid, also dissociates completely:

HCl (aq) → H+ (aq) + Cl- (aq)

The hydrogen ion from the acid then reacts with the sulfite ion to form sulfuric acid and water, creating a net ionic equation :

2H+ (aq) + SO3^2- (aq) → H2SO3 (aq)

Because of the solubility trends of sulfates and sulfites under standard conditions, the sulfuric acid produced also dissociates into ions:

H2SO3 (aq) → 2H+ (aq) + SO3^2- (aq)

Therefore, the overall net ionic equation is:

H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq)

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