A student dissolves 10.7 g of lithium chloride (LiCl) in 300. g of water in a well-insulated open cup. He then observes the temperature of the water rise from 22.0 °C to 28.6 °C over the course of 3.8 minutes. Use this data, and any information you need from the ALEKS Data resource:
LiCl(s) rightarrow Li+(aq) + Cl-(aq)
You can make any reasonable assumptions about the physical properties of the solution. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction.
1) Is this reaction exothermic, endothermic, or neither?
2) If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in this case.
3) Calculate the reaction enthalpy delta Hrxn per mole of LiCl.

Answers

Answer 1

Answer:

1) Exothermic.

2) [tex]Q_{rxn}=-8580J[/tex]

3) [tex]\Delta _rH=-121.0kJ/mol[/tex]

Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature increases the reaction is exothermic because it is releasing heat to solution, that is why the temperature goes from 22.0 °C to 28.6 °C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is absorbed by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

[tex]Q_{rxn}=-m_{Total}C(T_2-T_1)\\\\Q_{rxn}=-(300g+10.7g)*4.184 \frac{J}{g\°C} (28.6\°C-22.0\°C)\\\\Q_{rxn}=-8580J[/tex]

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case LiCl, we proceed as follows:

[tex]\Delta _rH=\frac{Q_{rxn}}{n_{LiCl}} \\\\\Delta _rH=\frac{-8580J}{10.7g*\frac{1mol}{150.91g} }*\frac{1kJ}{1000J} \\\\\Delta _rH=-121.0kJ/mol[/tex]

Best regards!


Related Questions

A student dissolves 15.0 g of ammonium chloride(NH4Cl) in 250. 0 g of water in a well-insulated open cup. She then observes the temperature of the water fall from 20.0 oC to 16.0 oC over the course of minutes. Use this data, and any information you need from the ALEKS Data resource, to answer the questions below about this reaction:
NH4Cl(s) rightarrow NH4+(aq) + Cl-(aq)
You can make any reasonable assumptions about the physical properties of the solution. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction.
1. Is this reaction exothermic, endothermic, or neither?
2. If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in this case.
3. Calculate the reaction enthalpy deltaHrxn per mole of NH4CI.

Answers

Answer:  

1) Endothermic.  

2) [tex]Q_{rxn}=4435.04J[/tex]  

3) [tex]\Delta _rH=15.8kJ/mol[/tex]

Explanation:  

Hello there!  

1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.  

2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

[tex]Q_{rxn}=-(15.0g+250.0g)*4.184\frac{J}{g\°C}(16.0-20.0)\°C\\\\ Q_{rxn}=4435.04J[/tex]    

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:

[tex]\Delta _rH=\frac{ Q_{rxn}}{n}\\\\\Delta _rH= \frac{ 4435.04J}{15.0g*\frac{1mol}{53.49g} } *\frac{1kJ}{1000J} \\\\\Delta _rH=15.8kJ/mol[/tex]

Best regards!  

Best regards!

Calculate the molality of a solution containing 15.0 g of ethylene glycol (C2H6O2) dissolved in 145 g of water.

Answers

Answer:  

Molarity=moles of solute/ L of solution

Molality = moles of solute/ kg of solvent

Solute= what is being dissolved

Solvent= what is doing the dissolving

Solution= both together

Explanation:

Example's:

#1.  For number one you use the Molarity formula.  M= moles of solute/ L of solution.

To find moles of Mg(NO3)2 you divide 95g by its molar mass which is 148.33g so 95/148.33=.6405 moles of Mg(NO3)2.  Then plug in what you have.  .38M= .6405 moles Mg(NO3)2 / X.  Then solve for X using algebra.  .6405/.38= 1.686 L of solution.  (Volume).

Final Answer: 1.686 L

#2.  For number 2 you use the Molality formula.  m= moles of solute/ kg of solvent.

First you have to find moles of glucose by taking 267g and dividing it by its molar mass which is 180.56g.               267g/180.56g= 1.532 moles of glucose.  Then you have to change L to kg.  The easiest way to do this is to look at the density and see that for every 1 ml there is 1 gram.  So to take Liters to ml you multiply 1.59 by 1000 and get 1590 ml.  So that means you have 1590 grams. then you divide 1590grams by 1000 to get 1.59 Kg of slovent.  Then plug in your information into the formula.      molality= 1.532 moles of glucose / 1.59 Kg of solvent= .964 molality.

Final answer: .964 mol/Kg

#3.  m= moles of solute / Kg of solvent.  0.445 mol solute / 2.07 Kg solvent= .215 Molality

Final Answer: .215 mol/Kg

#4. m= moles of solute / Kg of solvent.  take 13.5g and divide it by ethylene glycols molar mass which is 62.068 g.  13.5g / 62.068g= .218 mol.  Then you take 135g of water and divide it by 1000 to get Kg.  135/1000=.135 Kg.  Then plug in your information.  m= .218mol/.135 Kg= 1.615 molality

Final Answer: 1.615 mol/Kg.

Why isn't salt water safe for humans to drink?
O A. Salt water has too many water molecules and too few salt molecules.
B. Salt water has too few water molecules and too many salt molecules.
O
C. Salt water contains molecules that are poisonous to humans.
O
D. Salt water contains molecules that are too large for humans to process.

Answers

Answer:

B. Salt water has too few water molecules and too many salt molecules

Explanation:

Seawater is toxic to humans because your body is unable to get rid of the salt that comes from seawater. Your body normally gets rid of excess salt by having the kidneys produce urine, but it needs freshwater to dilute the salt in your body for the kidneys to work properly.

What type of intermolecular forces occur between 2 oil molecules?

Answers

Answer:

The three main kinds of intermolecular interactions are dipole-dipole interactions, London dispersion forces, and hydrogen bonds

Explanation:

If an equilibrium mixture of the three gases at 600K contains 2.92*10^-2 M COCH(g) and 1.76*10^2 M CO, what is the equilibrium
concentration of Cl2?

Answers

Answer:

C

Explanation:

A neutron is a negatively-charged particle in the atom. true or false

Answers

Answer: true

Explanation:

why phenol is more acid than alcohol????​

Answers

Answer:

Phenol is more acidic than alcohol due to resonance stabilization of the phenoxide ion.

How many mL of 0.715 M HCl is required to neutralize 1.25 grams of sodium carbonate? (producing carbonic acid)
I really couldn't find the answer since molarity and volume for sodium carbonate are not given.
I will mark the correct answer with steps as best answer.

Answers

Answer:

34 mL

Explanation:

We'll begin by calculating the number of mole in 1.25 g of sodium carbonate, Na₂CO₃. This can be obtained as follow:

Mass of Na₂CO₃ = 1.25 g

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)

= 46 + 12 + 48

= 106 g/mol

Mole of Na₂CO₃ =?

Mole = mass /molar mass

Mole of Na₂CO₃ = 1.25 / 106

Mole of Na₂CO₃ = 0.012 mole

Next, we shall determine the number of mole HCl needed to react with 0.012 mole of Na₂CO₃.

The equation for the reaction is given below:

Na₂CO₃ + 2HCl —> H₂CO₃ + 2NaCl

From the balanced equation above,

1 mole of Na₂CO₃ reacted with 2 moles of HCl.

Therefore, 0.012 mole of Na₂CO₃ will react with = 0.012 × 2 = 0.024 mole of HCl.

Next, we shall determine the volume of HCl required for the reaction. This is illustrated:

Mole of HCl = 0.024 mole

Molarity of HCl = 0.715 M

Volume of HCl =?

Molarity = mole /Volume

0.715 = 0.024 / volume of HCl

Cross multiply

0.715 × volume of HCl = 0.024

Divide both side by 0.715

Volume of HCl = 0.024 / 0.715

Volume of HCl = 0.034 L

Finally, we shall convert 0.034 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.034 L = 0.034 L × 1000 mL / 1 L

0.034 L = 34 mL

Therefore, 34 mL of HCl is needed for the reaction.

The amount of HCl required for counterbalancing 1.25 g of Na2CO3(Sodium Carbonate) would be:

- [tex]34 ml[/tex]

Given that,

Mass of Na2CO3 [tex]= 1.25 g[/tex]

To find the Moles of Na2CO3, we will find the molar mass of Na2CO3,

Molar Mass of or Na2CO3 [tex]= 106 g/mol[/tex]

So,

Moles of Na2CO3 [tex]= mass /molar mass[/tex]

[tex]= 1.25/106[/tex]

[tex]= 0.012 mol[/tex]

To determine the quantity of HCl required to display the reaction with 0.012 mol of Na2CO3

[tex]Na_{2} CO_{2} + 2HCl[/tex][tex]H_{2}CO_{3} + 2NaCl[/tex]

While balancing the equation, we know that [tex]0.012[/tex] × [tex]2 = 0.024 mole of HCl[/tex] is necessary to process the reaction.

Now,

As we know,

HCl moles [tex]= 0.024[/tex]

Molarity of HCl [tex]= 0.715 M[/tex]

∵ Quantity of HCl required = Moles/Molarity

[tex]= 0.024 / 0.715[/tex]

[tex]= 0.034 l[/tex] [tex]or 34ml[/tex]

Thus, 34 ml is the correct answer.

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Help meee??
HNO 3 is an example of Select one :
Triprotic Acid
O Diprotic Acid
O Monoprotic Acid
Tetraprotic Acid​

Answers

Answer: Monoprotic Acid

If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the
rate of production of nitrogen and hydrogen? Given 2NH3 3H2 + N2

Answers

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

The Answer is: 3.15 × 10⁻⁶ mol H₂/L.s; 1.05 × 10⁻⁶ mol N₂/L.s

Step 1:  When we Write the balanced equation

Then 2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

After that, The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:Now 2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

After that, The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:Now 2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

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Dominant traits are more common in a population then recessive traits
false
true

Answers

Answer:

No Just because a trait is dominant does not mean that it is present in the population

Explanation:

Nitrous oxide (N2O), more commonly known as laughing gas, is used as a mild sedatitive during various dental procedures.As a gas, it has a densityof 1.977 x 10-3g/mL.Wheniron is exposed to oxygen it forms rust (Fe2O3), which is a solid and has a density value of 5.25 g/mL.Why are the density values so different among these substances?
a)The metal atoms weigh more than the atoms of the gas.
b)The metal forms metallic bonds which are more greatly affected by gravity, increasing the mass.
c)The metal is a solid, and solids weigh more based on the principles of their states of matter.
d)There are fewer gas particles than solid particles in the same volume.

Answers

Answer:

B.

Explanation:

The metal forms metallic bonds which are more greatly affected by gravity, increasing the mass.

WILL GIVE THE BRAINLIEST!!! help me pls

Answers

Answer:

Two of them are solids, one is liquid. Two of them are edible, one is not. One is a mixture, and two are not.

Explanation:

Lipids include:
A. fats and water
B. Oils and carbohydrates
C. Waxes and sterols

Answers

I’m pretty sure it’s A

Answer: C

waxes and sterols

Explanation:

edge 2021

How many molecules of O2 will be formed from 0.500 grams of KCIO3

Answers

Answer:

3.7 x 10²¹ O₂ molecules

Explanation:

2KClO₃ => 2KCl + 3O₂

given 0.500g KClO₃ => 0.500g/122.55g/mol = 0.0041 mole KClO₃

0.0041 mole KClO₃ => 3/2(0.0041) mole O₂ = 0.0061 mole O₂

0.0061 mole O₂ = 0.0061 mole O₂ x 6.02 x 10²³ molecules O₂/mole O₂

= 3.7 x 10²¹ molecules O₂

About how many atoms of Mg are present in 24.3 grams of Mg?
5.85 x 1025
4.35 x 1019
c
4.81 X 1024
6.02 x 103

Answers

Answer:

D is the answer i think

Explanation:

how many ions does magnesium phosphite have? how many of those are anions? please explain thought process.

Answers

Answer:

We can simply refer to the cation in the ionic compound as magnesium. Phosphorus, Pstart text, P, end text, is a group 15 element and therefore forms 3- anions.

Explanation:

Because it is an anion, we add the suffix -ide to its name to get phosphide as the name of the ion.

Walking up a flight of stairs and noticing that it gets warmer as you climb
higher is an example of *
5 points
Conduction
Convection
Radiation
Brain damage

Answers

I think the answer is C Radiation

Which of the following methods of fossil formation is formed in dry areas due to a lack of moisture?

a
Mummification
b
Casts and molds
c
Amber fossil
d
Carbonization

Answers

I don’t know the answer but my guess would be mummification

The methods of fossil formation in dry areas due to a lack of moisture is mummification. The correct option is a.

What is mummification?

Mummification is the process of deliberately drying or embalming flesh to preserve the body after death.

This typically entailed removing moisture from a deceased body and desiccating the flesh and organs with chemicals or natural preservatives such as resin.

Mummification served the purpose of keeping the body intact so that it could be transported to a spiritual afterlife.

The body was stripped, positioned on a slanted table, and washed in natron solution (a naturally occurring salt used both as soap and a preservative).

The brain was extracted from the skull through a hole in the ethmoid bone (the bone separating the nasal cavity from the skull cavity).

Thus, the correct option is A.

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A gas is initially at 20.0 o C. That temperature is
K.

The gas is then heated to 30.0 0 C. That temperature is
K.


The initial pressure of the gas is 1.00 atm. The new pressure of the gas is
atm. (Round to 2 decimal places)

Answers

Answer:

The temperature of the gas is 293.15 K.

The temperature of the gas is 303.15 K.

The new pressure of the gas is 0.97 atm.

Explanation:

You need to add 273.15 to the Celsius temperature to convert from Celsius to Kelvin.

So,

20.0°C + 273.15 = 293.15 K.

30.0°C + 273.15 = 303.15 K.

We can use the following equation to calculate the new pressure:

[tex]P_2 = P_1 *\frac{T_2}{T_1}[/tex]

where:

P1 is the initial pressure P2 is the final pressureT1 is the initial temperatureT2 is the final temperature

Plugging in the values from the problem, we get:

[tex]P_2 = 1.00 atm * \frac{303.15 K }{293.15 K} = 0.97 atm[/tex]

Therefore, the new pressure of the gas is 0.97 atm.

What is a biome?
Question 1 options:

An area with lots of of rainfall.


An area with the same climate, landscape, plants, and animals.


An area with different temperatures.


A bunch of trees, forests, and rivers.

Answers

Answer:

An area with the same climate, landscape, plants, and animals.

An area with landscape, plants, climate, animals

How many moles are contained in 2.3 liters of a 1.2M solution?

Answers

Answer:

[tex]\boxed {\boxed {\sf 2.76 \ mol}}[/tex]

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

[tex]molarity = \frac {moles}{liters}[/tex]

We know the molarity is 1.2 M (mol\liter) and there are 2.3 liters of solution. Substitute the known values into the formula.

[tex]1.2 \ mol/liter= \frac {x}{2.3 \ liters}[/tex]

Since we are solving for x, we must isolate the variable. It is being divided by 2.3 and the inverse of division is multiplication. Multiply both sides by 2.3 liters.

[tex]2.3 \ liters *1.2 \ mol/liter= \frac {x}{2.3 \ liters}* 2.3 \ liters\\2.3 *1.2 \ mol= x\\2.76 \ mol =x[/tex]

In a solution with a molarity of 1.2 and 2.3 liters of solution, there are 2.76 moles.

The local weather station is predicting that a warm front will pass through Charlotte, NC in two days.
How will Charlotte's weather most likely be affected by the passing of a warm front?

A The wind speed will decrease.
B The air pressure will increase.
C the temperature will increase.
D The temperature will decrease.

Answers

Answer:

C the temperature will increase.

Which sublevels hold valence electrons?

Answers

Answer:

The Periodic Table can be divided into s, d, p, and f sublevel blocks. For elements in the s sublevel block, all valence electrons are found in s orbitals. For elements in the p sublevel block, the highest energy valence electrons are found in p orbitals.

Explanation:

How many milliliters of a 25% (m/v) NaOH solution would contain 75 g of NaOH?
A 19 mL
B) 25 mL
C
33 mL
D
75 mL
E
3.0 x 102 mL

Answers

Answer:

E  3.0 x 10² mL.

Explanation:

Hello there!

In this case, according to the formula for the calculation of the mass-volume percent:

[tex]\% m/V=\frac{m_{solute}}{V_{solution}}*100\%[/tex]

Whereas it is necessary to know the mass of the solute and the volume of the solution. Thus, given the mass of NaOH as the solute, the volume of the solution would be:

[tex]V_{solution}=\frac{m_{solute}}{\% m/V}*100\%[/tex]

Then, by plugging in we obtain:

[tex]V_{solution}=\frac{75g}{25\%}*100\%\\\\V_{solution}=3.0x10^2mL[/tex]

Thus, the answer is E  3.0 x 10² mL.

Best regards!

how do you think a device could change the sound that we hear? Make sure you use vocabulary such as frequency, energy and amplitude

Answers

Answer:

good luck tho

Explanation:


How many significant figures are in the number 420,000 mi?

Answers

Answer:

2

Explanation:

zeros are not considered sig figs

Which best describes the law of conservation of mass? 0 The coefficients in front of the chemicals in the reactants should be based on the physical state of the products. O Products in the form of gases are not considered a part of the total mass change from reactants to products. O When reactants contain both a solid and a liquid, the solid counts toward the overall mass and the liquid does not. O The mass of the reactants and products is equal and is not dependent on the physical state of the substances.​

Answers

Explanation:

pdrias darme la traduccion no te entiendo

A compound containing nitrogen and oxygen is decomposed in the laboratory. It produces 24.5 g nitrogen and 70.0 g oxygen. Calculate the empirical formula of the compound.

Answers

Answer:

N2O5

Explanation:

1. Convert to moles

24.5g N * 1mol/14g = 1.75

70.0g * 1mol/16g = 4.375

2. Divide each value by the smallest

1.75/1.75 = 1

4.375/1.75 = 2.5

3. Multiply each by a whole number so that they are both whole numbers

1*2 = 2

2.5*2 = 5

4. These are moles of elements present in the compound

Answer: N2O5

Question: You decompose a compound containing nitrogen and oxygen in the laboratory and produce 24.5 g of nitrogen and 70.0 g of oxygen. Calculate the empirical formula of the compound.

True or False. The scientific method is a rigid process?

Answers

The answer to your question is true

The statement "The scientific method is a rigid process" is false.

What is scientific method ?

The method by which scientist search for solutions and answers to their problem and question with the help of procedure is called scientific method.

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