1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
seropon [69]
3 years ago
7

A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution? Useful formulas..

Chemistry
2 answers:
max2010maxim [7]3 years ago
7 0
They are asking you for OH- ions concentration, which is why you have to find the pOH of the solution. You can easily do this with this formula:

pH + pOH = 14

Solve for pOH and you get:

pOH = 14 - pH

Replace the pH with it's value:

pOH = 14 - 11.8
pOH = 2.2

Now that you have the pOH you can find the concentration of OH- ions with this formula:

pOH = -log[OH-]

Solve the formula for [OH-] and you get:

[OH-] = 10^(-pOH)

Now replace pOH with it's value:

[OH-] = 10^(-2.2)
[OH-] = 0.006309 M

The concentration of OH- atoms is 0.006309M or in the scientific notation 6.309×10^(-3)M
olya-2409 [2.1K]3 years ago
3 0

Answer:

Its B

Explanation:

Took the test on E D G E

You might be interested in
How can You friend electrons, protons and neutrons in an element?
Charra [1.4K]
You cant its not really that possible there diffrent elements to gather them into one source would be difficult more than likely a scientific explanation would answer this based upon test and research
7 0
3 years ago
what do Mars, Mercury, and Venus have in common? A. They have a gas surface composition and a thick atmosphere. b They have many
mario62 [17]

Answer:

the best guess would have to be C

3 0
3 years ago
Why can scientists ignore the forces of attraction between particles in a gas under ordinary conditions
Alika [10]
<span>Scientists ignore the forces of attraction between particles in a gas under ordinary conditions</span><span> because the particles in a gas are apart and moving fast, rather than clustered and moving slow, therefore the forces of attraction are too weak to have a visible effect.</span>
3 0
3 years ago
A
amm1812

Solution:

1) Separate out the half-reactions. The only issue is that there are three of them.

<span>Fe2+ ---> Fe3+ 
S2¯ ---> SO42¯ 
NO3¯ ---> NO</span>

How did I recognize there there were three equations? The basic answer is "by experience." The detailed answer is that I know the oxidation states of all the elements on EACH side of the original equation. By knowing this, I am able to determine that there were two oxidations (the Fe going +2 to +3 and the S going -2 to +6) with one reduction (the N going +5 to +2).

Notice that I also split the FeS apart rather than write one equation (with FeS on the left side). I did this for simplicity showing the three equations. I know to split the FeS apart because it has two "things" happening to it, in this case it is two oxidations.

Normally, FeS does not ionize, but I can get away with it here because I will recombine the Fe2+ with the S2¯ in the final answer. If I do everything right, I'll get a one-to-one ratio of Fe2+ to S2¯ in the final answer.

2) Balancing all half-reactions in the normal manner.

<span>Fe2+ ---> Fe3+ + e¯ 
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯ 
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O</span>

3) Equalize the electrons on each side of the half-reactions. Please note that the first two half-reactions (both oxidations) total up to nine electrons. Consequently, a factor of three is needed for the third equation, the only one shown below:

<span>3 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]</span>

Adding up the three equations will be left as an exercise for the reader. With the FeS put back together, the sum of all the coefficients (including any that are one) in the correct answer is 15.

Problem #2: CrI3 + Cl2 ---> CrO42¯ + IO4¯ + Cl¯ [basic sol.]

Solution:

Go to this video for the solution

Problem #3: Sb2S3 + Na2CO3 + C ---> Sb + Na2S + CO

Solution:

1) Remove all the spectator ions:

<span>Sb26+ + CO32- + C ---> Sb + CO</span>

Notice that I did not write Sb3+. I did this to keep the correct ratio of Sb as reactant and product. It also turns out that it will have a benefit when I select factors to multiply through some of the half-reactions. I didn't realize that until after the solution was done.

2) Separate into half-reactions:

<span>Sb26+ ---> Sb 
CO32- ---> CO 
C ---> CO</span>

3) Balance as if in acidic solution:

<span>6e¯ + Sb26+ ---> 2Sb 
2e¯ + 4H+ + CO32- ---> CO + 2H2O 
H2O + C ---> CO + 2H+ + 2e¯Could you balance in basic? I suppose, but why?</span>

4) Use a factor of three on the second half-reaction and a factor of six on the third.

<span>6e¯ + Sb26+ ---> 2Sb 
3 [2e¯ + 4H+ + CO32- ---> CO + 2H2O] 
6 [H2O + C ---> CO + 2H+ + 2e¯]The key is to think of 12 and its factors (1, 2, 3, 4, 6). You need to make the electrons equal on both sides (and there are 12 on each side when the half-reactions are added together). You get 12 H+ on each side (3 x 4 in the second and 6 x 2 in the third). You get six waters with 3 x 2 in the second and 6 x 1 in the third.Everything that needs to cancel gets canceled!</span>

5) The answer (with spectator ions added back in):

<span>Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO</span>

6) Here's a slightly different take on the solution just presented.

<span>a) Write the net ionic equation:<span>Sb26+ + CO32- + C ---> Sb + CO</span>b) Notice that charges must be balanced and that we have zero charge on the right. So, do this:<span>Sb26+ + 3CO32- + C ---> Sb + CO</span>c) Now, balance for atoms:<span>Sb26+ + 3CO32- + 6C ---> 2Sb + 9CO</span>d) Add back the sodium ions and sulfide ions to recover the molecular equation.<span>Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO</span></span>

7) Here's a discussion of a wrong answer to the above problem.

However, after reading the above wrong answer example, look at problem #10 below for an instance of having to add in a substance not included in the original reaction.

Problem #4: CrI3 + H2O2 ---> CrO42¯ + IO4¯ [basic sol.]

Solution:

1) write the half-reactions:

<span>Cr3+ ---> CrO42¯ 
I33¯ ---> IO4¯ 
H2O2 ---> H2O</span>

I wrote the iodide as I33¯ to make it easier to recombine it with the chromium ion at the end of the problem.

2) Balance as if in acidic solution:

<span>4H2O + Cr3+ ---> CrO42¯ + 8H+ + 3e¯ 
12H2O + I33¯ ---> 3IO4¯ + 24H+ + 24e¯ 
2e¯ + 2H+ + H2O2 ---> 2H2O</span>

I used water as the product for the hydrogen peroxide half-reaction because that gave me a half-reaction in acid solution. It will all go back to basic at the end of the problem.

3) Recover CrI3 by combining the first two half-reactions from just above:

<span>16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯</span>

4) Equalize the electrons:

<span>2 [16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯] 
27 [2e¯ + 2H+ + H2O2 ---> 2H2O]leads to:32H2O + 2CrI3 ---> 6IO4¯ + 2CrO42¯ + 64H+ + 54e¯ 
54e¯ + 54H+ + 27H2O2 ---> 54H2O</span>

5) Add the half-reactions together. Strike out (1) electrons, (2) hydrogen ion and (3) water. The result:

<span>2CrI3 + 27H2O2 ---> 2CrO42¯ + 6IO4¯ + 10H+ + 22H2O</span>

6) Add 10 hydroxides to each side. This makes 10 more waters on the right, so combine with the water alreadyon the right-hand side to make 32:

<span>2CrI3 + 27H2O2 + 10OH¯ ---> 2CrO42¯ + 6IO4¯ + 32H2O</span>



3 0
3 years ago
Explain the difference between cocl2 6h2o and anhydrous cobalt chloride
lana66690 [7]
<span>Colbat (ii) which is a compound birth out of the combination of chlorine and colbat to form Cocl2.6h2o has water in it as we can see from it's chemical it's hexahydrate Anhydrous cobalt chloride as the word anhydrous clearly states , does not have water in</span>
8 0
4 years ago
Other questions:
  • PLEASE HELP ASAP!!!
    6·2 answers
  • Write a net ionic equation for the overall reaction that occurs when aqueous solutions of potassium hydroxide and phosphoric aci
    5·1 answer
  • How does a community look after a volcano erupts
    12·2 answers
  • ANSWER FIRST GET BRAINLY
    5·1 answer
  • The degree to which various compounds will dissociate in solution varies greatly.
    13·1 answer
  • Differentiate between a precipitate and an aqueous solution
    7·1 answer
  • Give an example to illustrate the difference between "paying more in taxes ” and paying a higher tax rate .
    14·1 answer
  • Is backbonding is possible in BF4- ? If no then why?
    8·1 answer
  • Which of these features was most likely formed by a divergent boundary?
    12·2 answers
  • Which color???????<br> please help
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!