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nikdorinn [45]
3 years ago
7

A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing point depression of

the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
7 0
From the equation; ΔTf = Kf × m
Where, Kf for water = 1.853 K kg/mole; m is the molarity = number of solute/amount of solvent in kg.
Glucose is the solute whose molecular mass is 180 g/mole and water is the solvent. 
Moles of solute = 15.5/180 = 0.0861 moles
Amount of solvent in kg = 245/1000 = 0.245 Kg
Therefore; molarity = 0.0861/0.245 = 0.3515 moles/Kg
Therefore; ΔTf = 1.853 × 0.3515 = 0.6513 K
Hence; the depression in freezing point is 0.6513 
The freezing point of solution will therefore be;
= 273 - 0.6513 = 272.3487 K

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Answer:

Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.

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How much heat must be removed from 456 g of water at 25. 0 c to hcange it into ice at 10 c
ValentinkaMS [17]

The heat required to convert water is given as the product of the mas and the latent heat. 28.34 kJ heat must be removed from the water to change it into ice.

<h3>What is heat energy?</h3>

Heat energy is the product of mass, specific heat capacity, and temperature change. It is given as,

\rm Q = mc \Delta T

Given,

Mass of water = 456 gm

Specific heat capacity = 4.186 J / g K

Temperatutre change = 14.85 K

Substituting values above:

\begin{aligned} \rm Q &= 456 \times 4.186 \times 14.85\\\\&= 28345.91\;\rm J\\\\&= 28.34\;\rm kJ\end{aligned}

Therefore, 28.34 kJ of heat energy should be removed from the water.

Learn more about the heat here:

brainly.com/question/14052023

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6 0
2 years ago
Identify the following as physical (P) or chemical (C) changes.
Stels [109]

holacomo es tu pregunta nola entiendo

lanation:

4 0
3 years ago
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¿How do the products of the reaction to the phenol red test and the splint test? Please help me it's for today.! :((​
laila [671]

Answer:

The answer in this question is show you made Sodium Hydroxide and Hydrogen Gas.In order to do the products of the reaction relate to the phenol red test and the splint test you need to show that you made Sodium Hydroxide and Hydrogen Gas. Show that you made Sodium Hydroxide and Hydrogen Gas so that the products of the reaction relate to the phenol red test and the splint test.

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How are the following aspects of a reaction affected by the addition of a catalyst? 1) activation energy of the reverse reaction
snow_tiger [21]

These are four questions and four answers.

Answers:

1) activation energy of the reverse reaction

     b. Decreased

2) Rate of the forward reaction

    a. Increased

3) Rate of the reverse reaction

    a. Increased

4) Activation energy of the forward reaction

    b. decreased

Explanation:

<em>Activarion energy</em> is the energy required by the reactants to form the intermediate transition state and become products.

<em>Catalysts</em> are substances that change the path of the chemical reactions, lowering the activation energy, and thus speeding up the rate of the reactions, since the products can reach the new lower activation energy faster.

The equilibrium reactions are the chemical process in which two reactions, the <em>forward and the reverese reactions</em>, occur simultaneously and at the same rate.  The equlibrium reactions may be represented by:

  • A ⇄ B

Where A → B is the direct or forward reaction, and A ← B is the reverse reaction (note the inversed arrow, from right to left).

For the direct reaction A represents the reactants and B represents the products. On the other hand, B represents the reactants and A represents the reactants of the reverse reaction and A. This, is A is the reactant of the forward reaction and the product of the reverse reaction, while B is the reactant of the reverse reaction and the product of the forward reaction.

Since, <em>the addition of a catalyst</em> lowers the activation energy of the process, the new activation energy is lower for both the forward and the reverse reaction, meaning that:

1. <em>The activation energy of the reverse reaction is decreased</em> (option b. of the first question)

2.<em> The rate of the forward reaction is increased</em> (option a. of the second question)

3. <em>The rate of the reverse reaction is increased</em> (option a. of the third question).

4. <em>Activation energy of the forward reaction is decreased</em> (option b. of the fourth question).

In summary, the addition of a catalyst decreases the activation energy for both forward and reverse reactions, and increases the rate of both forward and reverse reactions.

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