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kaheart [24]
3 years ago
12

What is the concentration of O2(g), in parts per million, in a solution that contains 0.008 gram of O2(g) dissolved in 1000. gra

ms of H2O(l)?
Chemistry
2 answers:
AnnyKZ [126]3 years ago
5 0

concentration of O₂ dissolved in H₂O can be written as ppm

ppm stands for parts per million - mg/kg

the amount of mg in 1 kg of solution

the mass of O₂ - 0.008 g

1000 mg equivalent to 1 g

therefore mass of O₂ in mg - 0.008 g x 1000 mg/g = 8 mg

the mass of water is 1000 g

1000 g is equivalent to 1 kg

mass of water in kg - 1000 g / 1000 g/kg = 1 kg

there's 8 mg of O₂ in 1 kg of water

therefore concentration of O₂ is - 8 mg/kg

also can be written as 8 ppm

answer is 8 ppm

gregori [183]3 years ago
4 0

Answer : The concentration of O2(g) in parts per million is, 8 ppm

Solution : Given,

Mass of oxygen gas (solute) = 0.008 g

Mass of water (solvent) = 1000 g

First we have to calculate the mass of solution.

Mass of solution = Mass of solute + Mass of solvent = 0.008 + 1000 = 1000.008 g

Now we have to calculate the concentration of O2(g) in parts per million.

ppm : It is defined as the mass of solute present in one million (10^6) parts by mass of the solution.

ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6

Now put all the given values in this expression, we get

ppm=\frac{0.008g}{1000.008g}\times 10^6=7.99=8ppm

Therefore, the concentration of O2(g) in parts per million is, 8 ppm

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A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati
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Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

\text{calculating the initial HI}= \frac{mol}{V}

                                       =\frac{9.3 \times 10 ^ -3}{2}

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\text{Similarly}\ \  I_2 \ \  \text{follows} \ \  H_2 = 0 }

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

HI  = 0.00465 - 2x\\\\ I_{2}  \ eq = H_2 \ eq = 0 + x \\\\

It is defined that:

I_2 = 6.29 \times 10^{-4}  \ M \\\\x = I_2 \\\\

HI \ eq= 0.00465 - 2x \\

          =0.00465 -2 \times 6.29 \times 10^{-4} \\\\ =  0.00465 -\frac{25.16 }{10^4}  \\\\   = 0.003392\  M

Now, we calculate the position:  

For the reaction H 2(g) + I 2(g)\rightleftharpoons  2HI(g), you can calculate the value of Kc at 1000 K.  

data expression for Kc

2HI \rightleftharpoons  H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}

         = \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386

calculating the reverse reaction

H_2(g) + I_2(g)\rightleftharpoons  2HI(g)

Kc = \frac{1}{Kc} \\\\

     = \frac{1}{0.034386}\\ \\= 29.081\\

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gregori [183]
Hello!

After the addition of a small amount of acid, a reasonable value of buffer pH would be 5,15.

If initially there are equal amounts of a weak acid and its conjugate base, the pH would be equal to the pKa, according to the Henderson-Hasselbach equation:

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