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Y_Kistochka [10]
3 years ago
8

For Kw, the product of [H3O+] and [OH-] is ___________.

Chemistry
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

Kw = 1×10⁻¹⁴

Explanation:

The product between the [H₃O⁺] .  [OH⁻] talks about the

autoionization of water.

2H₂O  ⇄  H₃O⁺  .  OH⁻     Kw

This is an equilibrium of 2 molecules of water that produces 1 mol of hydronium and 1 mol of hydroxides.

In order to determine pH and pOH, parameters to indicate acidic or basic solutions we know:

- log [H₃O⁺] = pH

- log [OH⁻] = pOH

When [H₃O⁺] or [OH⁻] = 1×10⁻⁷ , pH or pOH = 7 (Neutral solutions)

So if we see the equation above, we have an equilibrium.

In order to determine Kw, we need [H₃O⁺] .  [OH⁻]

As [H₃O⁺] and [OH⁻] = 1×10⁻⁷ , Kw = 1×10⁻¹⁴

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4 0
3 years ago
What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
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Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

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As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

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PV = nRT

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V = volume

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Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

4 0
3 years ago
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