You are in a laboratory creating a new chocolate bar. You want to create the sweetest chocolate bar by maximizing the sugar concentration. You are doing this by adding the sugar to a chocolate mixture. Which would allow you to dissolve more sugar?

The answer: Add the sugar after heating the mixture.

Answers

Answer 1

Adding the sugar after heating the mixture would allow you to dissolve more sugar, which would result in a sweeter chocolate bar.

When you dissolve sugar in a liquid, such as in a chocolate mixture, there is a limit to the amount of sugar that can be dissolved at a given temperature. This limit is known as the solubility of the sugar in that liquid. The solubility of sugar in water is higher at higher temperatures, which means that you can dissolve more sugar in hot water than in cold water. The same principle applies to chocolate mixtures.

By heating the chocolate mixture, you increase the temperature of the mixture, which in turn increases the solubility of the sugar in the mixture. This allows you to dissolve more sugar in the mixture than if you were to add the sugar to the mixture at room temperature or when it is cold.

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--The given question is incorrect, the correct question is

"You are in a laboratory creating a new chocolate bar. You want to create the sweetest chocolate bar by maximizing the sugar concentration. You are doing this by adding the sugar to a chocolate mixture. Which would allow you to dissolve more sugar?"--

Related Questions

2
Select the correct answer.
Which phrase best describes heat?
OA.
B.
OC.
D.
the energy that an object has as a result of its temperature
the average translational kinetic energy of the particles in an object
the energy transferred between objects at different temperatures
the total amount of energy possessed by the particles in an object

Answers

Heat is most accurately described as "the energy transferred between objects at different temperatures" (C). Until they reach thermal equilibrium, or the same temperature, heat is a type of energy that flows freely from a hotter to a colder item.

Heat can be transferred through conduction, convection, or radiation. The temperature differential between the items and the thermal conductivity of the materials involved determine how much heat is transported.

Temperature, a measurement of the average kinetic energy of the particles in an item, is not the same as heat. Internal energy is the entire amount of energy held by an object's particles, which includes both their kinetic and potential energies.

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1. HCI (aq) + NaOH (aq) → NaCl (aq) + H₂O (1)
a. What are the reactants?
b. What are the products?

Answers

In the given reaction, the reactants are hydrochloride acid (HCI) and sodium hydroxide (NaOH). The products are sodium chloride (NaCl) and water.

The chemical equation provided represents a neutralization reaction between hydrochloric acid (HCI) and sodium hydroxide (NaOH) in aqueous solution.

A neutralization reaction is a type of double displacement reaction in which an acid and a base react to form salt and water.

The reactants in this equation are hydrochloric acid (HCI) and sodium hydroxide (NaOH). Hydrochloric acid is a strong acid that dissociates in water to form hydrogen ions (H+) and chloride ions (Cl-). Sodium hydroxide, on the other hand, is a strong base that dissociates in water to form sodium ions (Na+) and hydroxide ions (OH-).

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help w calorimeter problems pls.

Answers

1. The specific heat capacity of the metal is 0.102 J/gºC

2. The specific heat capacity of the metal is 0.432 J/gºC

3. The final temperature of water is 16.7 °C

1. How do I determine the specific heat capacity of the metal?

First, we shall obtain the heat absorbed by the water. This is shown below:

Volume of water = 125 mLMass of water (M) = 125 gInitial temperature (T₁) = 22 °CFinal temperature (T₂) = 25.4 °CChange in temperature (ΔT) = 25.4 - 22 = 3.4 °CSpecific heat capacity of water (C) = 4.184 J/gºC Heat absorbed by water (Q) =?

Q = MCΔT

Q = 125 × 4.184 × 3.4

Q = 1778.2 J

Finally, we shall determine the specific heat capacity of the metal. This is shown below:

Heat absorbed by water (Q) = 1778.2 JHeat released by metal (Q) = -1778.2 JMass of metal (M) = 2.36×10² gInitial temperature (T₁) = 99.5 °CFinal temperature (T₂) = 25.4 °CChange in temperature (ΔT) = 25.4 - 99.5 = -74.1 °CSpecific heat capacity of metal (C) = ?

Q = MCΔT

-1778.2 = 2.36×10² × C × -74.1

-1778.2 = -17487.6 × C

Divide both sides by -17487.6

C = -1778.2 / -17487.6

Specific heat capacity of metal = 0.102 J/gºC

2. How do I determine the specific heat capacity of the metal?

As discussed above, we shall first obtain the heat absorbed by the water. This is shown below:

Volume of water = 75.2 mLMass of water (M) = 75.2 gInitial temperature (T₁) = 20.5 °CFinal temperature (T₂) = 28.6 °CChange in temperature (ΔT) = 28.6 - 20.5 = 8.1 °CSpecific heat capacity of water (C) = 4.184 J/gºC Heat absorbed by water (Q) =?

Q = MCΔT

Q = 75.2 × 4.184 × 8.1

Q = 2548.56 J

Finally, we shall determine the specific heat capacity of the metal. This is shown below:

Heat absorbed by water (Q) = 2548.56 JHeat released by metal (Q) = -2548.56 JMass of metal (M) = 95.3 gInitial temperature (T₁) = 90.5 °CFinal temperature (T₂) = 28.6 °CChange in temperature (ΔT) = 28.6 - 90.5 = -61.9 °CSpecific heat capacity of metal (C) = ?

Q = MCΔT

-2548.56 = 95.3 × C × -61.9

-2548.56 = -5899.07 × C

Divide both sides by -5899.07

C = -2548.56 / -5899.07

Specific heat capacity of metal = 0.432 J/gºC

3. How do i determine the final temperature of water?

The final temperature is the same as the equilibrium temperature of the mixture. Thus, we shall obtain the equilibrium temperature. Details below:

Mass of warm water (Mᵥᵥ) = 100Temperature of warm water (Tᵥᵥ) = 50 °CMass of cold water (M) = 50 gTemperature of cold water (T) = 20 °CEquilibrium temperature (Tₑ) =?

Heat loss by warm water = Heat gain by cold

MᵥᵥC(Tᵥᵥ - Tₑ) = MC(Tₑ - T)

Cancel out C

Mᵥᵥ(Tᵥᵥ - Tₑ) = M(Tₑ - T)

100 × (50 - Tₑ) = 50 × (Tₑ - 20)

Clear bracket

1500 - 100Tₑ = 50Tₑ - 1000

Collect like terms

1500 + 1000 = 50Tₑ + 100Tₑ

2500 = 150ₑ

Divide both side by 150

Tₑ = 2500 / 150

Tₑ = 16.7 °C

The equilibrium temperature is 16.7 °C.

Thus, we can conclude that the final temperature of the water is 16.7 °C

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Similar to For Practice 14.8) Determine the freezing point of an aqueous solution that contains 0.867 m glycerin (CHgOz).
Ki(water) = 1.86°C/m and Kg(water) - 0.512°C/m. Freezing point of water = 0.0 °C.

Similar to For Practice 14.3) Find the mass (in grams) of glucose (CH1206) in 505 mL of 10.5% glucose solution by mass. Assume the density of the solution is 1.04g/mL

Answers

The freezing point of the solution containing 0.867 m glycerin is -1.61442 °C. Option C is correct

The mass of glucose in 505 mL of 10.5% glucose solution is 53.01 g or 5.30 x 10^2 g.

Option C is correct

To find the freezing point depression of the solution containing 0.867 m glycerin:

ΔTf = Kf * molality

ΔTf = (1.86°C/m) * 0.867 m

ΔTf = 1.61442 °C

The freezing point depression is 1.61442 °C.

The freezing point of the solution is:

Freezing point = 0.0 °C - ΔTf

Freezing point = 0.0 °C - 1.61442 °C

Freezing point = -1.61442 °C

To find the mass of glucose in 505 mL of 10.5% glucose solution:

Mass of glucose = Volume of solution * Density of solution * % mass

Mass of glucose = 505 mL * 1.04 g/mL * 10.5%

Mass of glucose = 53.01 g

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