All categories

3 years ago

What is the pH of a 9.50×10−2 M ammonia solution??

1 answer:

4 0

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

c=9,50*10⁻² mol/L

Ammonia solution is a weak electrolyte.
The dissociation constant of ammonium hydroxide is:
K=1.78*10⁻⁵

[OH⁻]=√(Kc)

pH=14-pOH=14+lg[OH]⁻
pH=14+lg(√(Kc))

pH=14+lg(√(1.78*10⁻⁵*9.50*10⁻²))=8.23
pH=8.23

You might be interested in

An atom has the following electron configuration 1s2 2s2 2p6 3s2 3p4 . How many valence electrons does this Atom have

yawa3891 [41]

Answer:

Explanation:

This atom is sulfur (if the electrons are equal to the protons/not an ion). You can tell the number of valence electrons by looking at the individual shell. The first shell (1s) can only hold 2 electrons. The second shell (2s and 2p) can hold 8 electrons. The third shell (3s and 3p), which is the valence shell, only has 6 out of its possible 8 electrons, so this atom has 6 valence electrons.

6 0

3 years ago

Scientific explanations are not always based on empirical evidence <br><br> True or false?

tigry1 [53]

True some explanations are not always based on empirical evidence

4 0

3 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago

Elaborate on the difference in natural occurrences between fission and fusion reactions. A) Neither fission nor fusion reactions

vazorg [7]

<span>Okay then I would go with choice B since fusion takes place in the sun which is a giant star.</span>

8 0

3 years ago

Read 2 more answers

Jim drops an iron magnet during a chemistry lab experiment. He finds that it is no longer magnetic. What has happened in terms o

jok3333 [9.3K]

Answer: option A.

A physical change occurred in which the iron remained iron, but lost one of its physical properties

Explanation: magnetization and demagnetization is a physical change as no new substance is form and it is easily reversed. When an iron rod is magnetized, it can loss it magnetic property by hitting it. So when an iron rod which is magnetized falls from a height it can also loss its magnetic property

4 0

3 years ago

Read 2 more answers