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2 years ago

What is the pH of 2.0 M nitrous acid at equilibrium (Ka= 4.6×10-4)?

2 answers:

7 0

To find pH, you need this formula: pH= -log [H+]

as you can see, we need to find first the concentration of H+.

you find it, we can use the ka value. Ka is the dissociation constant for acid and it can be used to determine the H+ concentration

Ka= (x)^2/ initial concentration

x = [H+], so all we need to do is solve for X.

ka= 4.6x10-4
initial= 2.0M

so now the formula looks like this:

4.6x10-4= (x)^2/ 2.0

x=0.0303 = [H+]

now that we have the H+ concentration, we can plug it in the pH formula to get our answer

pH= -log (0.0303)= 1.52

Serga [27]2 years ago

5 0

Check out this link to help you. :)
Someone else already answered this question.
brainly.com/question/1489637

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In living things what is the source of the carbon-14 that is used in radiocarbon dating

Mumz [18]

In living things, the source of the carbon-14 that is used in radiocarbon dating is carbon dioxide in the atmosphere.
Living things inhale oxygen, and exhale carbon dioxide into the atmosphere, which is why the air and atmosphere are so full of it.

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3 years ago

Read 2 more answers

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

The decomposition of a compound at 400⁰C is first order with half-life of 1570 seconds. what fraction of an initial amount of th

kirill115 [55]

Answer: After 4710 seconds, 1/8 of the compound will be left

Explanation:

Using the formulae

Nt/No = (1/2)^t/t1/2

Where

N= amount of the compound present at time t

No= amount of compound present at time t=0

t= time taken for N molecules of the compound to remain = 4710 seconds

t1/2 = half-life of compound = 1570 seconds

Plugging in the values, we have

Nt/No = (1/2)^(4710s/1570s)

Nt/No = (1/2)^3

Nt/No= 1/8

Therefore after 4710 seconds, 1/8 molecules of the compound will be left

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2 years ago

How many atoms of carbon are in 24 grams of carbon?

Mamont248 [21]

(3 grams of carbon) x (1 mole of carbon/12 grams ) =3/12 = 1/4 of a mole of carbon. Then... ( 1/4 of a mole) x (6.02 x 10^23 atoms/mole) = approximately 1.5 x 10^23 atoms.

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3 years ago

In a reaction vessel of 10.0 I, heat 9.00 g of water so that it partially decomposes. In the equilibrium mixture, 0.250 mol of w

MatroZZZ [7]

Answer:

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Explanation:

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3 years ago