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How many formula units are in 5.33 moles of cucl2

Blizzard [7]

Answer:

32.1 × 10²³ formula units of CuCl₂

Explanation:

Given data:

Number of moles of CuCl₂ = 5.33 mol

Number of formula units of CuCl₂ = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

one mole of any substance contain 6.022 × 10²³ formula units thus,

5.33 moles of CuCl₂ = 5.33 ×6.022 × 10²³ formula units

32.1 × 10²³ formula units of CuCl₂

3 0

3 years ago

Metallic pan is provided with a wooden plastic handle give reason

AlladinOne [14]

The metallic pan iis most likely going to be used on a stove.

The stove is heating something, and the conductive metallic pan will, well, conduct that heat throughout the entire body of the pan. Doing this will spread the heat to the handle, burning your hands.

Both wood and plastic are insulators, and they do not conduct heat or electricity. They will insulate your hands and protect them from the heat.

5 0

2 years ago

Why do you think the molarity relationship is so important to the chemist?

Oksana_A [137]

There are two big advantages of using molarity to express concentration. The first advantage is that it's easy and convenient to use because the solute may be measured in grams, converted into moles, and mixed with a volume.

The second advantage is that the sum of the molar concentrations is the total molar concentration. This permits calculations of density and ionic strength

5 0

2 years ago

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

2 years ago

How many particles are in 697.008 grams of Potassium Sulfate?

g100num [7]

Answer:

174.2592

Explanation:

Trust me.

8 0

2 years ago