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S_A_V [24]
3 years ago
10

What is the hydrogen ion concentration in a 0.10M solution of KOH?​

Chemistry
1 answer:
dybincka [34]3 years ago
7 0

KOH is a strong base, so [OH-] = 0.10 M = 1.0 x 10^-1 M

[H+][OH-] = Kw

Kw = 1.0 x 10^-14

[H+] = 1.0 x 10^-14 / 1.0 x 10^-1 = 1.0 x 10^-13 M

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6 0
4 years ago
Do metals gain or lose electrons as they form ions
Anna71 [15]

Answer:

They usually lose electrons to form ions with 2 positive charges

Explanation:

6 0
3 years ago
Read 2 more answers
the elements iron and oxygen can react to form the compound iron oxide. what type of reaction is this? please explain.
rjkz [21]
Your reaction
.. Fe + O2 ---> FexOy
for this reaction..
.. the Fe on the left is in the 0 oxidation state
.. the Fe on the right is in the +(2y/x) oxidation state
.. the O on the left is in the 0 oxidation state
.. the O on the right is in the -2 oxidation state
meaning
.. the O is reduced... . . (it's reduced in oxidation state)
.. the Fe is oxidized.. . .(oxidation state increased)
this is a REDOX reaction

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AND.. it's also a synthesis reaction.. (aka combination reaction)
4 0
3 years ago
Read 2 more answers
mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of
anygoal [31]

<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

  • <u>For nitrogen gas:</u>

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

  • <u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0
3 years ago
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