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3 years ago

How many atoms are in 3.20 moles of carbon

Chemistry

1 answer:

DIA [1.3K]3 years ago

6 0

Answer:

19.264× $10^{23}$ atoms are present in 3.2 moles of carbon.

Explanation:

It is known that one mole of each element is composed of Avagadro's number of atoms. This is same for all the elements in the periodic table.

So, as 1 mole of any element = Avagadro's number of atoms = 6.02× $10^{23}$ atoms

It is as simple as understanding a dozen of anything is equal to 12 pieces of that object.

As here the moles of carbon is given as 3.20 moles, the number of atoms in this mole can be determined as below.

1 mole of carbon = 6.02 × $10^{23}$ atoms

Then, 3.20 moles of carbon = 3.20 × 6.02 × $10^{23}$ atoms

Thus, 19.264× $10^{23}$ atoms are present in 3.2 moles of carbon.

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Arrange the following aqueous solutions in order of increasing boiling point temperature (lowest to highest temperature, with 1

Elanso [62]

Answer:

0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.

Explanation:

Step 1: Data given

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = Shows how much the boiling point increases

⇒i = the van't Hoff factor: Says in how many particles the compound will dissociate

⇒ Since all are aqueous solutions Kb for all solutions is the same (0.512 °C/m)

⇒m = the molality

Step 2:

0.20 m glucose

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for glucose = 1

⇒ Kb = 0.512 °C/m

⇒m = 0.20 m

ΔT = 1*0.512 * 0.20

ΔT = 0.1024 °C

0.30 m BaCl2

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for BaCl2 = Ba^2+ + 2Cl- : i = 3

⇒ Kb = 0.512 °C/m

⇒m = 0.30 m

ΔT = 3*0.512 * 0.30

ΔT = 0.4608 °C

0.40 m NaCl

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for NaCl = Na+ + Cl- : i = 2

⇒ Kb = 0.512 °C/m

⇒m = 0.40 m

ΔT = 2*0.512 * 0.40

ΔT = 0.4096 °C

0.50 m Na2SO4.

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for Na2SO4 = 2Na+ + SO4^2- : i =3

⇒ Kb = 0.512 °C/m

⇒m = 0.50 m

ΔT = 3*0.512 * 0.50

ΔT = 0.768 °C

0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.

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