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3 years ago

8

Determine the Ka for CH3NH3+ at 25*C. The Kb for CH3NH2 is 4.4 x 10-4.

1 answer:

vfiekz [6]3 years ago

8 0

Answer:

Ka= 2.3 x 10^-11

Explanation:

so Ka x Kb = Kw where Kw=10^-14 at 25 degrees Celsius

plugging in Kb and Kw, we have

Ka x 4.4 x 10^-4 = 10^-14

solving for Ka,

Ka= (10^-14)/(4.4 x 10^-4)

Ka= 2.3 x 10^-11

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Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

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$V_1=23.3~mL$

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I hope it helps!

8 0

3 years ago

4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below

gayaneshka [121]

Answer : The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

$q_{rxn}=-q_{cal}$

$q_{cal}=c_{cal}\times \Delta T$

where,

$q_{rxn}$ = heat released by the reaction = ?

$q_{cal}$ = heat absorbed by the calorimeter

$c_{cal}$ = specific heat of calorimeter = $97.1kJ/^oC=97100J/^oC$

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=(27.282-25.000)=2.282^oC$

Now put all the given values in the above formula, we get:

$q_{cal}=(97100J/^oC)\times (2.282^oC)$

$q_{cal}=221582.2J=221.6kJ$

As, $q_{rxn}=-q_{cal}$

So, $q_{rxn}=-221.6kJ$

Thus, the heat of the reaction is -221.6 kJ

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