Please answer this question

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Answer 1

Answer:

Answer:

I'm not an expert at this, but I assume its mercury.

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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:

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:

$m_(PbI_2)=18.2gPbI_2$

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.

What is the molarity of .00000093 grams of epinephrine in one liter?

Answers

Answer:

5.076 * 10 ⁻¹² M

Explanation:

molarity of a solution is given as -

Molarity (M) = ( w / m ) / V ( in L)

where ,

m = molecular mass ,

w = given mass ,

V = volume of solution ,

From the question ,

w = given mass of epinephrine = 0.00000093 g

V = 1 L

Since , the molecular mass of epinephrine = m = 183,204 g/mol

M = ?

In order to determine the molarity of epinephrine , the above formula is used , by putting the respected values ,

Molarity (M) = ( w / m ) / V ( in L)

M = ( 0.00000093 g / 183,204 g/mol ) / 1 L

M = 5.076 * 10 ⁻¹² M

The chemical equation for this reaction is Ca + O2→CaO. What is the product, or what are the products, of this reaction?

Answers

Answer: Cao2

Explanation: the reason for this is because the chemical reactions on both sides should be equal and since the O has a 2 the O should have 2 on the other equation.

In 100 grams of sweet peas there are 14.5 carbohydrates,5.7 grams of sugars,5.1 grams of dietary fiber, 5.4 grams of protein and 0.4 grams of fat what is the percent composition of each?

Answers

14.5 % carb
5.7% sugar
5.1% fiber
5.4% protein
0.4% fat

Solve the problems. Express your answers to the correct number of significant figures. 0.09 + 1.324 =
41.0 + 78.3 =

Answers

Ans: 1) 1.41

2) 119.3

While adding two numbers, the rule of significant figures requires that the number of decimal places in the final answer is equal to that of the term with the least decimal places

In the given examples:

1) 0.09 - has 2 decimal places

1.324- has 3 decimal places

Sum = 0.09 + 1.324 = 1.414

Round off to 2 decimal places = 1.41

2) 41.0 - has 1 decimal place

78.3- has 1 decimal place

Sum = 41.0 + 78.3 = 119.3

Final answer will also have 1 decimal place = 119.3

1. 1.141

2. 119.3

HOPE THIS HELPS.

How many joules of heat energy are absorbed when 80.0 g of water are heated from 10.0°C to 50.0°C? *

Answers

Answer:

13440 J

Explanation:

c ≈ 4200 J / (kg * °C)

m = 80 g = 0,08 kg

$t_(1)$ = 10 °C

$t_(2)$ = 50 °C

The formula is: Q = c * m * ( $t_(2) - t_(1)$ )

Calculating:

Q = 4200 * 0,08 * (50 - 10) = 13440 (J)