All categories

2 years ago

How many atoms are in 4.5 x 1022 moles of CO2?

Chemistry

1 answer:

weeeeeb [17]2 years ago

8 0

The number of atoms : N = 2.709 x 10⁴⁶

<h3>Further explanation</h3>

Given

4.5 x 10²² moles of CO₂

Required

The number of atoms

Solution

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

Input the value :

N = 4.5 x 10²² x 6.02 x 10²³

N = 2.709 x 10⁴⁶

You might be interested in

The study of the composition of matter is?

NemiM [27]

Chemistry. More specifically, analytical chemistry.

6 0

3 years ago

Characterize EACH of the three given statements as being TRUE or FALSE and then indicate the collective true-false status of the

Aleks04 [339]

Answer:

A = True

B = False

C = True.

I think this is the answer

8 0

2 years ago

If the solubility of KCl in 100 mL of H₂O is 34 g at 20 °C and 43 g at 50 °C, label each of the following solutions as unsaturat

sertanlavr [38]

Answer:

a) Unsaturated

b) Supersaturated

c) Unsaturated

Explanation:

A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.

An unsaturated solution contains less solute than it has the capacity to dissolve.

A supersaturated solution, contains more solute than is present in a saturated solution. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.

According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at 20 °C is 34 g, and at 50 °C is 43 g we can label the solutions:

a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated

b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated

c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution with no precipitate ⇒ unsaturated (if it were saturated it would have had precipitate)

8 0

3 years ago

What group is the element NITROGEN in? 1 Group 4 2 Group 5 3 Group 2 4 Group 8

Butoxors [25]

Answer:

r7ñopdyiwriduwyzgpxpo

3 0

2 years ago

Read 2 more answers

One mole of a monatomic ideal gas is subjected to the following sequence of steps: a. Starting at 300 K and 10 atm, the gas expa

Verdich [7]

Answer:

a) Q = 0; W = 0; ΔU = 0; ΔH = 0; ΔS = 0.09 atm.L/K

b) Q = 1250 J; W = 0; ΔU = 1250 J; ΔH = 1250 J; ΔS = -0.0235 atm.L/K

c) Q = 3653.545 J; W = - 3653.545 J; ΔU = 0; ΔH = 0; ΔS = - 3653.545 J

d) Q = - 2080 J; W = 830 J; ΔU = - 1250 J; ΔH = - 2080 J; ΔS = - 5.984 J/K

Explanation:

a) If there is a vacuum, the work is zero, as it is a free expansion, the volume increases, the pressure decreases, the temperature is constant and the internal energy is constant.

∴ n = 1 mole

∴ PV = RTn....ideal gas

∴ P1 = 10 atm

∴ R = 0.082 atm.L/K.mol

∴ T = 300 K = T2

∴ V2 = 3*V1

⇒ W = 0.....expands freely into vacuum

⇒ ΔU = Q = 0....first law

⇒ ΔS = - nR Ln(P2/P1).....ideal gas

∴ V1*P1/T1 = V2*P2/T2

∴ T1 = T2 = 300 K

⇒ P2 = V1*P1 / V2 = V1*P1 / 3V1 = 10 atm/3 = 3.33 atm

⇒ ΔS = - (1mol)*(0.082 atm.L/K.mol) Ln ( 3.33/10)

⇒ ΔS = 0.09 atm.L/K

∴ ΔH = ΔU + (P2V2 - P1V1) = 0 + 0 = 0

b) heated reversibly at constant volume:

⇒ W = 0 ...at constant volume

∴ T2 = 400 K; T1 = 300 K

∴ V1 = V2

⇒ Q = ΔU = CvΔT....first law

∴ Cv = 12.5 J/K.mol.....monoatomic ideal gas

∴ ΔT = 400 - 300 = 100 K

⇒ Q = ΔU = 12.5 J/mol.K * 100K = 1250 J/mol * 1 mol = 1250 J

∴ ΔH = ΔU + PΔV = ΔU + 0 = 1250 J

∴ ΔS = - nR Ln (P2/P1)

∴ P2/T2 = P1/T1...constant volume

∴ P1 = 3.33 atm

⇒ P2 = P1*T2 / T1 = (3.33 atm)*(400K) / (300K) = 4.44 atm

⇒ ΔS = - (1mol)*(0.082atm.L/K.mol) Ln (4.44/3.33)

⇒ ΔS = - 0.0235 atm.L/K

c) reversibly expanded at constant temperature:

∴ T1 = T2 = 400K

∴ V2 = 3*V1

∴ ΔU = 0...constant temperature

⇒ Q = - W....fisrt law

∴ W = - ∫ PdV..... reversibly expansion

∴ P = nRT/V... ideal gas

⇒ W = - nRT ∫ dV/V

⇒ W = - nRT Ln (V2/V1)

⇒ W = - (1mol)*(8.314 J/K.mol) Ln (3)

⇒ W = - 9.134 J/K *400K = - 3653.545 J

⇒ Q = - W = 3653.545 J

⇒ ΔH = ΔU + P1V1 - P2V2 = 0 + nRT1 - nRT2 = 0 + 0 = 0

∴ ΔS = - nR Ln(P2/P1)

∴ P1 = 4.44 atm

⇒ P2 = V1*P1*T2/ V2*T1 = V1*(4.44atm)*(400K) / (3.V1)*(400K)

⇒ P2 = 4.44atm/3 = 1.48 atm

⇒ ΔS = - (1mol)*(8.314 J/mol.K) Ln (1.48/4.44)

⇒ ΔS = -9.134J/K * 400K = - 3653.545 J

d) reversibly cooled at constant pressure:

∴ T2 = 300 K; T1 = 400 K

∴ P2 = P1

⇒ Q = ΔH = CpΔT

∴ Cp = 20.8 J/K.mol

∴ ΔT = 300 - 400 = - 100 K

⇒ Q = ΔH = 20.8 J/mol.K * ( -100K) = - 2080 J/mol * 1mol = - 2080 J

⇒ ΔU = nCvΔT = (1mol)*(12.5 J/mol.K)*( - 100K) = -1250 J

⇒ W = ΔU - Q = ΔU - ΔH = -1250 J - ( - 2080 J ) = 830 J

∴ ΔS = ∫ δQ/T = ∫ nCpdT/T

⇒ ΔS = nCp Ln (T2/T1)

⇒ ΔS = (1mol)*(20.8 J/mol.K) Ln (300/400) = - 5.984 J/K

7 0

3 years ago