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amm1812
3 years ago
5

Calculate the volume of "dry" hydrogen that would be produced by one mole of magnesium at

Chemistry
1 answer:
kakasveta [241]3 years ago
5 0

Answer:The volume of H2 = 243mL = 0.243 L

Pressure of H2 gas collected = 745 mm

Pressure of H2 dry gas = P of H2 collected - vapor pressure of water = 745- 23.78 = 721.22 mm

= 721.22mm /760mm/atm

Temperature = 25C = 25+273= 298 K

Using the ideal gas equation

PV = nRT

or PV = (mass/molar mass) RT

(721.22/760)atm x 0.243L = (mass/ 2g/mol) x 0.0821L.atm/mol.K x 298K

Thus mass of H2 formed = 0.0188 g

Explanation:

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2) The rate of the revere reaction increases.

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3) The concentration of product increases.

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4) The concentration of products decreases.

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What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe
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Answer:

a) pH = 4.213

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Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]

<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒  [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

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