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4 years ago

A 0.821 gram sample of pure NH F was treated with 25.0 mL of 1.00 M NaOH

Chemistry

1 answer:

Olegator [25]4 years ago

8 0

Answer:

0.0222 mole of NaOH is needed to react with NH4F

Explanation:

NH4F + NaOH --> NaF + NH3 + H2O

Data given

Mass of NH4F =0.821g, Concentration of NaOH= 1M, volume of NaoH =25ml

But mole = (CV)/1000

given mole of NaoH = (1 * 25)/1000 = 0.025moles of NaOH used

Molar mass of NH4F = 37g/mol

mole of NH4F used 0.821 / 37 = 0.0222 mole NH4F

Determine the excess and limiting reactant,

NaOH is in excess

0.025 - 0.0222 = 0.0028 mole NaOH excess

0.0222 mole of NaOH is required to react with NH4F

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Calculate 50g 25.0c to 95.000<br>O.897 J / (g.c)

chubhunter [2.5K]

Answer:

Q = 3139.5 j

Explanation:

Given data:

Mass = 50 g

Initial temperature = 25°C

Final temperature = 95°C

Specific heat capacity = 0.897 j/g.°C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

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c = specific heat capacity of substance

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3 years ago

If a pH of water is 5.3 then 4.3 is it 1.0 less acidic 1.0 less basic

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The pH of water is 7
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3 years ago

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Evgen [1.6K]

Answer:

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Explanation:

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3 years ago

What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?

Ymorist [56]

Answer:

$\% diss = 50\%$

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

$HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}$

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

$Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}$

Thus, it is possible to find x given the pH as shown below:

$x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M$

So that we can calculate the initial concentration of the acid:

$\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\$

$[HA]_0=3.64x10^{-5}M$

Therefore, the percent dissociation turns out to be:

$\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%$

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