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miss Akunina [59]
1 year ago
11

The next 3 questions will walk you through using the Henderson-Hasselbalch equation for the following question. For each step pr

ovide a numerical answer. A solution is made by combining 300mL of 0.020M trimethyl amine, that is (CH3)3N and a 500mL of a 0.03M solution of it's conjugate acid, that is (CH3)3NH . The Kb for trimethyl amine is 6.3x10-5. The Henderson-Hasselbalch equation uses pKa not Kb, Calculate the pKa....
Chemistry
1 answer:
kicyunya [14]1 year ago
8 0

Answer:

<em>The pKa is 13.0.</em>

Explanation:

pKa + pKb = 14

Given, Kb of trimethylamine = 6.3 × 10^{-5}

pKb = - log (6.3 × 10^{-5})

= 1.0

⇒ pKa = 14 - pKb = 14 - 1.0

<u>pKa = 13.0</u>

<em><u></u></em>

<em>Check: For most weak acids,  pKa ranges from 2 to 13.</em>

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Among the alkali metals, the tendency to react with other substances. A. does not vary among the members of the group. . B. incr
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Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t
mars1129 [50]

Answer:

Explanation:

N₂       + 3H₂     =     2 NH₃

1 vol                         2 vol

786 liters               1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }

V_2 =303.44 liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

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1 year ago
when you burn a log in the fireplace, the resulting ashes have a mass less than that of the original log. Account for the differ
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Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l
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Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol

Putting values in above equation, we get:

-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]

\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

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Empirical formula of a compound gives the proportions of the elements in that compound but it does not define the actual arrangement and number of atoms.

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Therefore, the empirical formula will be  CH_{2}O.

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