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Ede4ka [16]
3 years ago
5

Given the reaction: N2 + O2 = 2NO for which the Keq at 2273 K is 1.2 x 10-4

Chemistry
1 answer:
MAXImum [283]3 years ago
3 0

<u>Answer:</u>

(a): The expression of equilibrium constant is K_{eq}=\frac{[NO]^2}{[N_2][O_2]}

(b): The equation to solve the concentration of NO is [NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}

(c): The concentration of NO is 0.0017 M.

<u>Explanation:</u>

The equilibrium constant is defined as the ratio of the concentration of products to the concentration of reactants raised to the power of the stoichiometric coefficient of each. It is represented by the term K_{eq}

(a):

The given chemical equation follows:

N_2+O_2\rightarrow 2NO

The expression for equilbrium constant will be:

K_{eq}=\frac{[NO]^2}{[N_2][O_2]}

(b):

The equation to solve the concentration of NO follows:

[NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}            ......(1)

(c):

Given values:

K_{eq}=1.2\times 10^{-4}

[N_2]_{eq}=0.166M

[O_2]_{eq}=0.145M

Plugging values in equation 1, we get:

[NO]=\sqrt{(1.2\times 10^{-4})\times 0.166\times 0.145}

[NO]=\sqrt{2.88\times 10^{-6}}

[NO]=0.0017 M

Hence, the concentration of NO is 0.0017 M.

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Bad White [126]

76.88 I think im sorry if wrong

5 0
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Pure acetic acid (hc2h3o2) is a liquid and is known as glacial acetic acid. calculate the molarity of a solution prepared by dis
Alexandra [31]
In order to find the molarity of the solution, we first require the moles of acetic acid added. For this,we need the mass which is:

Mass = volume * density

Mass = 50 * 1.05
Mass = 52.5 grams


Moles = mass / molecular weight

Moles = 52.5 / 60.05 
Moles = 0.874 mol

Next, we know that the molarity of a solution is:

Molarity = moles / liter
Molarity = 0.874 / 0.5

Molarity = 1.75 M
3 0
3 years ago
Read 2 more answers
What are molecules, and how are the properties of molecules different from the atoms they come from? Give an example.
goldenfox [79]

Molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance.

While Atoms are single neutral particles,

Molecules are neutral particles made of two or more atoms bonded together.

Exaplmes for molecules

H2O (water)

N2 (nitrogen)

O3 (ozone)

CaO (calcium oxide)

C6H12O6 (glucose, a type of sugar)

NaCl (table salt

And examples for atoms

Neon (Ne)

Hydrogen (H)

Argon (Ar)

Iron (Fe)

Calcium (Ca)

Deuterium, an isotope of hydrogen that has one proton and one neutron.

Plutonium (Pu)

F-, a fluorine anion.

3 0
3 years ago
What is the molar out of a solution that contains 33.5g of CaCl2 in 600.0mL of water
omeli [17]

Answer:

Here's what I got.

Explanation:

Interestingly enough, I'm not getting

0.0341% w/v

either. Here's why.

Start by calculating the percent composition of chlorine,

Cl

, in calcium chloride, This will help you calculate the mass of chloride anions,

Cl

−

, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that

1

mole of calcium chloride contains

2

moles of chlorine atoms.

2

×

35.453

g mol

−

1

110.98

g mol

−

1

⋅

100

%

=

63.89% Cl

This means that for every

100 g

of calcium chloride, you get

63.89 g

of chlorine.

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every

100 g

of calcium chloride, you get

63.89 g

of chloride anions,

Cl

−

.

This implies that your sample contains

0.543

g CaCl

2

⋅

63.89 g Cl

−

100

g CaCl

2

=

0.3469 g Cl

−

Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

100 mL

of this solution.

Since you know that

500 mL

of solution contain

0.3469 g

of chloride anions, you can say that

100 mL

of solution will contain

100

mL solution

⋅

0.3469 g Cl

−

500

mL solution

=

0.06938 g Cl

−

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.

.

ALTERNATIVE APPROACH

Alternatively, you can start by calculating the number of moles of calcium chloride present in your sample

0.543

g

⋅

1 mole CaCl

2

110.98

g

=

0.004893 moles CaCl

2

To find the molarity of this solution, calculate the number of moles of calcium chloride present in

1 L

=

10

3

mL

of solution by using the fact that you have

0.004893

moles present in

500 mL

of solution.

10

3

mL solution

⋅

0.004893 moles CaCl

2

500

mL solution

=

0.009786 moles CaCl

2

You can thus say your solution has

[

CaCl

2

]

=

0.009786 mol L

−

1

Since every mole of calcium chloride delivers

2

moles of chloride anions to the solution, you can say that you have

[

Cl

−

]

=

2

⋅

0.009786 mol L

−

1

[

Cl

−

]

=

0.01957 mol L

−

This implies that

100 mL

of this solution will contain

100

mL solution

⋅

0.01957 moles Cl

−

10

3

mL solution

=

0.001957 moles Cl

−

Finally, to convert this to grams, use the molar mass of elemental chlorine

0.001957

moles Cl

−

⋅

35.453 g

1

mole Cl

−

=

0.06938 g Cl

−

Once again, you have

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

In reference to the explanation you provided, you have

0.341 g L

−

1

=

0.0341 g/100 mL

=

0.0341% m/v

because you have

1 L

=

10

3

mL

.

However, this solution does not contain

0.341 g

of chloride anions in

1 L

. Using

[

Cl

−

]

=

0.01957 mol L

−

1

you have

n

=

c

⋅

V

so

n

=

0.01957 mol

⋅

10

−

3

mL

−

1

⋅

500

mL

n

=

0.009785 moles

This is how many moles of chloride anions you have in

500 mL

of solution. Consequently,

100 mL

of solution will contain

100

mL solution

⋅

0.009785 moles Cl

−

500

mL solution

=

0.001957 moles Cl

−

So once again, you have

0.06938 g

of chloride anions in

100 mL

of solution, the equivalent of

0.069% m/v

.

Explanation:

i think this is it

8 0
2 years ago
The iron content of foods can be determined by dissolving them in acid (forming Fe3+), reducing the iron(III) to iron(II), and t
____ [38]

Answer:

  • <em>Oxidation half-reaction</em>:

Fe²⁺(aq) →  Fe³⁺(aq) + 1e⁻

  • <em>Reduction half-reaction</em>:

Ce⁴⁺(aq) + 1e⁻ →  Ce³⁺(aq)

Explanation:

The reaction that takes place is:

  • Fe²⁺(aq) + Ce⁴⁺(aq) → Fe³⁺(aq) + Ce³⁺(aq)

The <em>oxidation half-reaction</em> is:

  • Fe²⁺(aq) →  Fe³⁺(aq) + 1e⁻

It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.

The <em>reduction half-reaction</em> is:

  • Ce⁴⁺(aq) + 1e⁻ →  Ce³⁺(aq)

It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.

4 0
3 years ago
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