Is rubbing alcohol a mixture, compound or solutions?

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 is the percent composition by mass of sulfur in the compound MgSO4 (gram formula mass = 120 g/mol)?

slavikrds [6]

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate. Explanation: Molar mass of compound = 120 g/mol.

7 0

2 years ago

During ______ the composition of a substance does not change

nignag [31]

During the Physical changes

6 0

3 years ago

using the same balanced equation, what is the number of moles of carbon dioxide produced from 3.750 moles of propane?

kondor19780726 [428]

Answer:

11.25moles of CO2

Explanation:

First, let us generate a balanced equation for the reaction of propane to produce CO2. This reaction called Combustion. It is a reaction in which propane burns in air (O2) to produce CO2 and H20. The equation is given below:

C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation,

1mole of C3H8 produced 3moles of CO2.

Therefore, 3.750 moles of C3H8 will produce = 3.750 x 3 = 11.25moles of CO2

5 0

2 years ago

Give the number of lone pairs around the central atom and the molecular geometry of XeF4.A)0 lone pairs, tetrahedralB)1 lone pai

vovangra [49]

Answer: 2 lone pairs, square planar

Explanation:

Using the VSEPR ( Valence Shell Electron Pair Repulsion)Theory

To calculate the number of lone pairs electron can be done using the formula;

Number of electrons = ½ (V+N-C+A)

V mean valency of the central atom

N means number of monovalent bonding atoms

C means charge on cation

A means charges on anion

Therefore, to calculate the number of lone pair electron C=A=0;

Number of electrons = ½ (8+4) = 12/2 = 6

Number of bonding pair = 4

Number of lone pairs of electron = 6-4 = 2

The hybridrization of the compound is sp3d2 because the number of electrons around the central atom is 6.

The geometry of the compound is square planar and this is because of the repulsion between the bonding pair of electrons and lone pair of electrons which causes the lone pair of electrons to lie in a perpendicular plane in order to acquire stability.

4 0

3 years ago