Calculate the specific heat of the substance if 373 J is required to raise the temperature of a 312 gram sample by 25°C?

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

What rule/principle states that electrons in the same sublevel (p, d, or f) are placed in individual orbitals, before they are p

Radda [10]

Answer:

Hunds Rule

Explanation:

3 0

3 years ago

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How will the rate at which a solid solute dissolves change if the solution is stirred?

MakcuM [25]

Answer:

The rate at which the solute dissolves will increase.

Explanation:

If a solution is stirred, the rate at which a solute dissolves would increase substantially provided the solution is not yet saturated.

Stiring would cause more of the solution to come in contact with every part of the solute. It will increase the surface area of contact for the solution to act which will shoot up the rate of reaction. Stiring helps to bring solutes in solutions into a more close contact with the molecules or compounds of the medium.

6 0

3 years ago

A topographical map would be useful for which activity? A. planning a hiking trip B. studying plant growth over time C. studying

amm1812

The answer is A. planning a hiking trip

A topographical map would not help studying plant growth over time unless you are looking for a better altitude to plant said plants.

A topographical map would not help studying rainfall for one year unless your location was really so high or low that it affected your weather

And most of all, a topographical map would not be useful for planning a cuise across the Atlantic Ocean because the elevation of the sea is zero!

A topographical map would be useful for planning a hiking trip because there are many factors and details that a hike should have. Which includes height, distance, paths, and elevation.

5 0

3 years ago

How can you distinguish between crystalline allotropic modifications of Sulphur from those of amorphous allotrops?

Burka [1]

The crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

<h3>What is a crystalline substance?</h3>

A crystalline substance is one that has a definite arrangement of the atoms in the substance. An amorphous substance lacks this definite arrangement. We can see this arrangement when we conduct an X-ray crystallography of the sulfur.

Also, the crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

Learn more about sulfur:brainly.com/question/13469437

#SPJ1

4 0

2 years ago