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Natali5045456 [20]
3 years ago
10

A solution has a pH = 4.63. What is the [H3O+] concentration?

Chemistry
1 answer:
LUCKY_DIMON [66]3 years ago
7 0

Explanation:

The pH of a solution can you be found by using the formula

pH = - log [ { H_3O}^{+}]

Since we are finding the [H3O+] , substitute the value of the pH and find it's antilog

We have

4.63 =  -  log[ { H_3O}^{+}] \\ [ { H_3O}^{+}]   =  {10}^{ - 4.63}  \\   \\  = 2.344 \times  {10}^{ - 5} mol {dm}^{ - 3}

Hope this helps you

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Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar
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Answer:

The answer is "\bold{0.525\ \ atm^{-1}}"

Explanation:

Given equation:

2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)

Given value:

\Delta rH =-198.2 \ \ \frac{KJ}{mol}

Kp=1100 \ K

\Delta x = 2-(2+1)\\\\

     = 2-(2+1)\\\\= 2-(3)\\\\= -1

\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm)     &1-2x&0.5-x&2x\end{array}\right

calculating the total pressure on equilibrium=  (1-2x)+(0.5-x)+2x \ atm\\\\

                                                                         = 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\

\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to  x= 0.15\\\\

calculating the pressure in  So_2:

= (1-2 \times 0.15)

= 1-0.30 \\\\ =0.70 \ atm

calculating the pressure in  O_2:

= (0.5- 0.15)\\\\= 0.35 \ atm \\

calculating the pressure in  So_3:

= (2 \times 0.15)\\\\= (.30) \ atm \\\\

Calculating the Kp at 1100 K:

= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\=  0.5247 \ \  or \ \  0.525 \ \ atm^{-1}  \\\\

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Explanation:

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Explanation:

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Answer:

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ΔG can be:

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