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3 years ago

This is the last week of school for me and i need help pleaseee !!

Chemistry

1 answer:

viktelen [127]3 years ago

6 0

Answer:

The reaction in the cold pack is an endothermic reaction

Explanation:

An endothermic reaction is a reaction that absorbs energy (heat) from the surrounding to supply the required energy of activation that allows the occurrence of the reaction

Therefore, when a reaction is endothermic, it boundary or the system where the reaction is occurring feels cold to touch

Hence, the reaction in the cold pack is an endothermic reaction.

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When dissolving a solid or liquid, as the temperature of the solvent increases, the rate of dissolution?

Alinara [238K]

Answer:A. Increases

Explanation:

Heating or an increase in temperature increases the kinetic energy of particles thereby increasing their motion and how they relate and react with one another.

Increase in the temperature of the solvent is directly proportional to the rate of dissolution. The rate of dissolution increases due to the increase in kinetic energy. This makes the solvent particles interact faster with the solute particles thereby increasing the dissolution rate.

7 0

3 years ago

1. A 250g chunk of metal is heated with 400 joules of energy and the temperature goes from 20 °C to 25°C. What is its specific h

Paha777 [63]

The specific heat capacity of this chunk of metal is equal to 0.32 J/g°C.

<u>Given the following data:</u>

Mass of metal = 250g

Quantity of energy = 400 Joules

Initial temperature = 20°C

Final temperature = 20°C

To determine the specific heat capacity of this chunk of metal:

<h3>The formula for quantity of heat.</h3>

Mathematically, quantity of heat is given by the formula;

$Q = mc\theta$

<u>Where:</u>

Q represents the quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity.

∅ represents the change in temperature.

Making c the subject of formula, we have:

$c = \frac{Q}{m\theta}$

Substituting the given parameters into the formula, we have;

$c = \frac{400}{250 \times (25-20)}\\\\c = \frac{400}{250 \times 5}\\\\c = \frac{400}{1250 }$

Specific heat, c = 0.32 J/g°C.

Read more on specific heat here: brainly.com/question/2834175

5 0

2 years ago

When a neutral atom loses an electron, it forms an ion with a positive charge called a _________? Question 1 options:

Black_prince [1.1K]

It should be C) cation

8 0

3 years ago

Read 2 more answers

What is the difference between conjugate acid-base pair?

NeX [460]

Answer:

B. a H+ ion is the answer dear.

Explanation:

ㅗㅐㅔㄷ ㅅㅗㅑㄴ ㅗㄷㅣㅔ ㅛㅐㅕ.

5 0

3 years ago

At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago