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Ex1. how many moles correspond to 250 g of each of the following compounds?
a. NaCl
b. KNO,
c. H, 0,
d. KHSO,

Ex2. Calculate how many particles are contained in:
a. 1.5 g of potassium, K
b. 0.470 g of oxygen, 0,
c. 0.555 g of silver chloride,
AgCl

Ex3. Determines how many molecules of ethyl alcohol, C, H, OH (d - 0.79 g / mL), are in a bottle that contains 200 mL of it.

Respuesta :

ChBader

Answer:

Ex1 = 4.3 moles

Explanation:

for Ex 1

 ** Number of mole = Mass (g) / Molecular weight ' Mw'  (g/mole)

for a = NaCl , first you must be calculate Mw = look to periodic table

Atomic weight for Na is 22.9 = 23 g and for Cl is 35.45 = 35 g

so Mw for NaCl = 23+35 = 58 g/mole

so Number of Moles for NaCl 250 g is

Number of Moles = 250 / 58 = 4.3 moles

do same for others ...

good luck ..

Answer:

Ex1. a) n NaCl = 4.277 mol NaCl

b) n KNO2 = 2.938 mol KNO2

c) n H2O2 = 7.349 mol H2O2

d) n KHSO3 = 2.080 mol KHSO3

Ex2. a) 2.31 E22 particles contained in 1.5 g K

b) 1.77 E22 particles contained in 0.470 g O

c) 2.332 E21 particles contained in 0.555 g AgCl

Ex3. molecules C2H5OH = 2.065 E24 molecules contained in 200 mL

Explanation:

Ex1.   n = mass (m) / molecular weight (Mw)

a) Mw NaCl = 22.989 + 35.453 = 58.442 g/mol

⇒ n NaCl = (250 g NaCl)/(58.442 g/mol) = 4.277 mol NaCl

b) Mw KNO2 = 39.0983 + 14.0067 + (2)15.9994 = 85.104 g/mol

⇒ n KNO2 = (250 g KNO2)/(85.104 g/mol) = 2.938 mol KNO2

c) Mw H2O2 = (2)(1.00794) + (2)15.9994 = 34.015 g/mol

⇒ n H2O2 = (250 g H2O)/(34.015 g/mol) = 7.349 mol H2O

d) Mw KHSO3 = 39.0983 + 1.00794 + 32.065 + 15.9994(3) = 120.169 g/mol

⇒ n KHSO3 = (250 g KHSO3)/(120.169 g/mol) = 2.080 mol KHSO3

Ex2.  n =  (m) / (Mw)

∴ 1 mol ≡ 6.022 E 23 particles

a) Mw K = 39.0983 g/mol

⇒  K = (1.5 g K)×(mol/39.0983 g K)×(6.022 E23 part/mol) = 2.31 E22 particles

b) Mw O = 15.9994 g/mol

⇒ O = (0.470 g O)×(mol/15.9994 g O)×(6.022 E23 part/mol) = 1.77 E22 part

c) Mw AgCl = 107.8682 + 35.453 = 143.32 g/mol AgCl

⇒ AgCl = (0.555 g)×(mol/143.32 g)×(6.022 E23 part/mol) = 2.332 E21 part.

Ex3. ethyl alcohol ( C2H5OH)

∴ δ C2H5OH = 0.79 g/mL

∴ V = 200 mL

⇒ molecules C2H5OH = ?

  • 1 mol ≡ 6.022 E23 molecules

∴ Mw C2H5OH = (2)12.0107 + (5)1.00794 + 15.9994 + 1.00794 = 46.068 g/mol

⇒ molecules C2H5OH = (200 mL)×(0.79 g/mL)×(mol/46.068 g)×(6.022 E23 molecules/mol)

⇒ molecules C2H5OH = 2.065 E24 molecules C2H5OH

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