All categories

3 years ago

Calculate the pOH of a solution that contains 3.9 x 10-5 M H3O+ at 25°C.

Chemistry

1 answer:

exis [7]3 years ago

6 0

Answer:

Option b. 9.59

Explanation:

First, let us calculate the pH. This is illustrated below:

[H3O+] = 3.9 x 10-5 M

pH = —Log [H3O+]

pH = —Log [3.9 x 10-5]

pH = 4.41

Recall: pH + pOH = 14

4.41 + pOH = 14

Collect like terms

pOH = 14 — 4.41

pOH = 9.59

You might be interested in

How much of a 24-gram sample of Radium-226 will remain unchanged at the end of three half-life periods?

shutvik [7]

Answer:

The right answer is "3 g".

Explanation:

Given:

Initial mass substance,

$M_0=24 \ g$

By using the relation between half lives and amount of substances will be:

⇒ $M=\frac{M_0}{2^n}$

$=\frac{24}{2^3}$

$=3 \ g$

Thus, the above is the correct answer.

8 0

3 years ago

Perform the following operation

lukranit [14]

Answer:

(5.4 x 10³) x (1.2 x 10⁷) = 6.48 x 10¹⁰

With correct significant figures, the answer would be 6.5 x 10¹⁰.

4 0

2 years ago

List the following objects in order from lowest inertia to highest inertia

m_a_m_a [10]

I don’t see nun tho where’s the objects

5 0

2 years ago

Read 2 more answers

6. if a cross country runner travels at a speed of 10 mph and he runs for 2 hours, how many miles will he have run at the end of

storchak [24]

Explanation:

distance = speed x time

distance = 10 x 2

distance = 20miles

hope that helps :)

5 0

2 years ago

wo reactions and their equilibrium constants are given. A + 2 B − ⇀ ↽ − 2 C K 1 = 2.57 2 C − ⇀ ↽ − D K 2 = 0.226 A+2B↽−−⇀2CK1=2.

Papessa [141]

Answer: The value of equilibrium constant for the net reaction is 11.37

Explanation:

The given chemical equations follows:

Equation 1: $A+2B\xrightarrow[]{K_1} 2C$

Equation 2: $2C\xrightarrow[]{K_2} D$

The net equation follows:

$D\xrightarrow[]{K} A+2B$

As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K=K_1\times \frac{1}{K_2}$

We are given:

$K_1=2.57$

$K_2=0.226$

Putting values in above equation, we get:

$K=2.57\times \frac{1}{0.226}=11.37$

Hence, the value of equilibrium constant for the net reaction is 11.37

6 0

3 years ago