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3 years ago

Express your answer using two significant figures. 2.7 cm3 = m3 2.0 mm3= m3

Chemistry

1 answer:

natka813 [3]3 years ago

8 0

Answer:

2.7 cm³ = 2.7 x 10⁻⁶ m³

2.0 mm³ = 2.0 x 10⁻⁹ m³

Explanation:

The values are multiplied by the conversion ratio, being sure to raise the ratio to the power of 3 so that the units properly cancel:

(2.7 cm³)(1m/100cm)³ = 2.7 x 10⁻⁶ m³

(2.0 mm³)(1m/1000mm)³ = 2.0 x 10⁻⁹ m³

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4 years ago

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3 years ago

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M acetic acid (Ka = 1.75x10-5) with 0.150

mylen [45]

Answer:

a) pH = 2.793

b) pH = 4.280

c) pH = 4.933

d) pH = 8.816

e) pH = 8.861

f) pH = 8.891

Explanation:

a) VNaOH = 0 mL

∴ CH3COOH ↔ CHECOO- + H3O+

⇒ Ka = 1.75 E-5 = [ H3O+ ] * [ CH3COO-] / [ CH3COOH ]

mass balance:

⇒ C CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M

charge balance:

⇒ [ H3O+ ] = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 M - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.75 E-5 [ H3O+ ] - 2.625 E-6 = 0

⇒ [ H3O+ ] = 1.61146 E-3 M

⇒ pH = - Log [ H3O+ ] = 2.793

b) after 5.0 mL NaOH:

∴ CH3COOH + NaOH ↔ CH3COONa + H2O

⇒ C NaOH = (5 E-3 L * 0.150 mol/L) / (0.025+0.01 ) = 0.02143 M

⇒ C CH3COOH = ((0.025*0.150) - (0.01*0.150)) / (0.025 + 0.01) = 0.0643 M

mass balance:

⇒ 0.02143 + 0.0643 = [ CH3COOH ] + [ CH3COO- ] = 0.086 M

charge balance:

⇒ [ H3O+ ] + [Na ] = [ CH3COO- ]

⇒ [ H3O+ ] + 0.02143 = [ CH3COO- ]

⇒ Ka = [ H3O+ ] * ( [ H3O+ ] + 0.150 ) / (0.086 - 0.02143 - [ H3O+ ]) = 1.75 E-5

⇒ [ H3O+ ]² + 0.02143 [ H3O+ ] = 1.13 E-6 - 1.75 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.02144 [ H3O+ ] - 1.13 E-6 = 0

⇒ [ H3O+ ] = 5.26 E-5 M

⇒ pH = 4.28

c) after 15 mL NaOH:

⇒ C CH3COOH = 0.0375 M

⇒ C NaOH = 0.05625 M

mass balance:

⇒ 0.09375 M = [ CH3COO- ] +[ CH3COOH ]

charge balance:

⇒ [ H3O+ ] + 0.05625 = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ] * ([ H3O+ ] + 0.05625) / (0.09375 - 0.05625 - [H3O+])

⇒ [H3O+]² + 0.05625[H3O+] = 6.5625 E-7 - 1.75 E-5 [H3O+]

⇒ [ H3O+]² + 0.05626[H3O+] - 6.5625 E-7 = 0

⇒ [ H3O+ ] = 1.1662 E-5 M

⇒ pH = 4.933

d) after 25 mL NaOH:

⇒ C NaOH = 0.075 M

⇒ C CH3COOH = 0 M....equiv. point

⇒Kh = Kw/Ka = 1 E-14 / 1.75 E-8 = 5.7143 E-10 = [ OH-]² / ( 0.075 - [OH-])

⇒ [OH-]² + 5.7143 E-10[OH-] - 4.286 E-11 = 0

⇒ [ OH- ] = 6.5463 E-6 M

⇒ pOH = 5.184

⇒ pH = 8.816

e) after 40 mL NaOH:

⇒ C NaOH = 0.0923 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 5.275 E-11 = 0

⇒ [OH-] = 7.2624 E-6 M

⇒ pOH = 5.139

⇒ pH = 8.861

f) after 60 mL NaOH:

⇒ C NaOH = 0.106 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 6.05 E-11 = 0

⇒ [OH-] = 7.7782 E-6 M

⇒ pOH = 5.11

⇒ pH = 8.891

5 0

4 years ago