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Valentin [98]
3 years ago
13

The nonvolatile, nonelectrolyte testosterone, C19H28O2 (288.40 g/mol), is soluble in benzene C6H6. Calculate the osmotic pressur

e generated when 12.9 grams of testosterone are dissolved in 286 ml of a benzene solution at 298 K.
Chemistry
1 answer:
dolphi86 [110]3 years ago
6 0

Answer:

3.824 atm

Explanation:

From the ideal gas equation

P = mRT/MW × V

m is mass of testosterone = 12.9 g

R is gas constant = 82.057 cm^3.atm/mol.K

T is temperature of benzene solution = 298 K

MW is molecular weight of testosterone = 288.40 g/mol

V is volume of benzene solution = 286 ml = 286 cm^3

P = 12.9×82.057×298/288.4×286 = 3.824 atm

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Answer:

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Explanation:

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NaOH and bleach have several characteristics in common. They include all BUT one of the characteristics listed.
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Sodium hydroxide and bleach have several characteristics in common and that includes: both can conduct electricity in solution, both can be found in the cleaning products and they are able to produce hydroxide ions in solution. However, their pH are not less than 7. Sodium hydroxide has a pH of 13, while bleach has a pH of 11. Therefore, both of them has a pH greater than 7. So the characteristic that they don't have in common is letter A.
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3 years ago
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Answer:

a) pH = 13.176

b) pH = 13

c) pH = 12.574

d) pH = 7.0

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f) pH = 1.21

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HBr + NaOH ↔ NaBr + H2O

∴ equivalent point:

⇒ mol acid = mol base

⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)

⇒ Va = 0.025 L

a) before addition acid:

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⇒ <em>C </em>NaOH = 0.150 M

⇒ [ OH- ] = 0.150 M

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⇒ pOH = 0.824

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b) after addition 5mL HBr:

⇒ <em>C </em>NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M

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⇒ [ OH- ] = 0.1 M

⇒ pOH = 1

⇒ pH = 13

c) after addition 15mL HBr:

⇒ <em>C </em>NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M

⇒ <em>C </em>HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M

⇒ [ OH- ] = 0.0375 M

⇒ pOH = 1.426

⇒ pH = 12.574

d) after addition 25mL HBr:

equivalent point:

⇒ [ OH- ] = [ H3O+ ]

⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²

⇒ [ H3O+ ] = 1 E-7

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d) after addition 40mL HBr:

⇒ <em>C</em> HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M

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d) after addition 60mL HBr:

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