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3 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]3 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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Convert each of the measurements given to the new unit stated using the factor label method (dimensional analysis). Show all wor

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26 cm -0.26 m

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8 0

3 years ago

And non-flammable gases Noble gases are that have low chemical

gtnhenbr [62]

Answer:

Noble gases are odorless, colorless, and nonflammable gases that have low chemical reactivity.
The full valence electron shells of these atoms make noble gases extremely stable.
& they are unlikely to form chemical bonds because they have little tendency to gain or lose

electrons.

5 0

3 years ago

For the reaction 2NOCl(g) <-----> 2NO(g) + Cl2(g), Kc = 8.0 at a certain temperature. What concentration of NOCl must be p

Nat2105 [25]

Answer is: concentration of NOCl is 3.52 M.

Balanced chemical reaction: 2NOCl(g) ⇄ 2NO(g) + Cl₂(g).
Kc = 8.0.
[NOCl] = 1.00 M; equilibrium concentration.
[NO] = x.
[Cl₂] = x/2; equilibrium concentration of chlorine.
Kc = [Cl₂] ·[NO]² / [NOCl].
8.00 = x/2 · x² / 1.
x³/2 = 8.
x = ∛16.
x = 2.52 M.
co(NOCl) = [NOCl] + x.
co(NOCl) = 1.00 M + 2.52 M.
co(NOCl) = 3.52 M; the initial concentration of NOCl.

7 0

3 years ago

What is the molality of a solution of water and kcl if the freezing point of the solution is –3mc030-1.jpgc?

Natasha_Volkova [10]

We will use the expression for freezing point depression ∆Tf
∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is
∆Tf = 0C - (-3°C) = 3°C
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and one Cl- ion:
KCl → K+ + Cl-
Therefore, the molality m of the solution can be calculated as
3 = 2 * 1.86 * m
m = 3 / (2 * 1.86)
m = 0.80 molal

4 0

3 years ago

The following combinations are not allowed. If n and ml are correct, change the l value to create an allowable combination:

motikmotik

The allowable combination for the atomic orbital is n=3, l=1, $m_{l}$ =-1.

<h3>What are the three quantum numbers of an atomic orbital?</h3>

Three quantum numbers specify an atomic orbital:

- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.

- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),

- $\boldsymbol{m}_{l}$ is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus; $m_{l}$ values are integers from -l to =l(l limits ml)

For n = 3, l can have three values: 0, 1, and 2. Since $m_{l}$ values are integers from -l to 0 to +l, for l = 0 the value of $m_{l}$ cannot be -1 (l = 0 has $m_{l}$ = 0).

There are two l values that are consistent with n and $m_{l}$ values:

l=1 or 2

Therefore, the allowable condition is n=3, l=1, $m_{l}$ =-1.

To know more about quantum numbers, visit: brainly.com/question/16979660

#SPJ4

4 0

2 years ago