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Vedmedyk [2.9K]
2 years ago
13

he concentration of H2S in a saturated aqueous solution at room temperature is approximately 0.1 M. Ka1=8.9x10-8 Ka2=1.0x10-19 C

alculate the pH of the solution in 2 sig figs
Chemistry
1 answer:
nirvana33 [79]2 years ago
5 0

Answer:

PH = 4.0

Explanation:

We are given;

Ka1 = 8.9 × 10^(-8)

Ka2 = 1.0 × 10^(-19)

From the 2 values of K given above, we can see that Ka2 is far smaller than Ka1.

Thus, Positive hydrogen ion (H+) will be majorly formed from first dissociation which is Ka1.

Now, the breakdown of the H2S solution is;

H2S⇌[H+] + [HS−]

Thus;

Ka1 = [[H+] × [HS2^(-)]]/(H2S)

HS2^(-) also has a positive hydrogen ion.

Thus, we can rewrite as;

Ka1 = [[H+] × [H+]]/(H2S)

Ka1 = (H+)²/(H2S)

Concentration of H2S is given as 0.1M. Thus;

8.9 × 10^(-8) = (H+)²/0.1

(H+)² = 0.1 × 8.9 × 10^(-8)

(H+) = √(0.1 × 8.9 × 10^(-8))

(H+) = 0.00009433981

Now, PH is gotten from;

PH = -log (H+)

Thus;

PH = -log 0.00009433981

PH ≈ 4.0

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MNaHCO₃: 23+1+12+(48×3) = 84g
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.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........

84g ----- 60g
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So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
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B) Amount of CH3COOH is in excess.

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3 0
2 years ago
How many molecules of CBr4 are in 250 grams of CBr4
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Answer:- 4.54*10^2^3 molecules.

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It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of CBr_4  = 12+4(79.9)  = 331.6 g per mol

let's make the set up using dimensional analysis:

250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})

= 4.54*10^2^3 molecules

So, there will be 4.54*10^2^3 molecules in 250 grams of CBr_4 .


8 0
2 years ago
When uranium-235 atoms undergo fission, ________ is/are produced?
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6 0
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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t
Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?

We have :

Enthalpy changes of formation of following s:

\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol

\Delta H_{f,CuO}=-157.3 kJ/mol

\Delta H_{f,NO_2}= 33.2 kJ/mol

\Delta H_{f,O_2}= 0 kJ/mol (standard state)

\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]

The equation for the enthalpy change of the given reaction is:

\Delta H_{rxn} =

=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})

\Delta H_{rxn}=

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\Delta H_{rxn}=424 kJ

The enthalpy change for the given reaction is 424 kJ.

6 0
2 years ago
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