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Dominik [7]
2 years ago
9

Provide 3 examples of substances that are mixtures and list the substances in them.

Chemistry
2 answers:
adell [148]2 years ago
5 0
Bleach Sulfrate Meth and pure vinegar Salt Sodium Cloride
babunello [35]2 years ago
3 0
Salt and water
Sugar and salt
Salt and pepper
You might be interested in
How many moles of carbon in 6.64 moles of CCl2 F
Bad White [126]

Answer: 6.64 moles of carbon.

Explanation:

Given data:

Number  of moles of C = ?

Number of moles of CCl₂F₂ = 6.64 mol

Solution:

In one mole of CCl₂F₂ there is one mole of carbon two moles of chlorine and two moles of fluorine are present.

In 6.6 moles of CCl₂F₂ :

Moles of carbon = 6.64 × 1 = 6.64 moles of carbon.

Moles of chlorine = 6.64× 2 = 13.28 moles of chlorine

Moles of fluorine = 6.64× 2 = 13.28 moles of fluorine

Read more on Brainly.com - brainly.com/question/15602143#readmore

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
What is acid according to arhenias concept?
nignag [31]

Answer:

Acid are those substances which release H + ions when dissolved in water.

Get that hundooo!

7 0
3 years ago
What volume of 0.152 M KMnO4 solution would completely react with 20.0 mL of 0.381 M FeSO4 solution according to the following n
ANEK [815]

<u>Answer:</u> The volume of permanganate ion (potassium permanganate) is 10.0 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of ferrous sulfate solution = 0.381 M

Volume of solution = 20.0 mL = 0.020 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.381M=\frac{\text{Moles of ferrous sulfate}}{0.020L}\\\\\text{Moles of ferrous sulfate}=(0.381mol/L\times 0.020L)=0.00762mol

For the given chemical equation:

5Fe^{2+}+8H^++MnO_4^-\rightarrow 5Fe^{3+}+Mn^{2+}+4H_2O

By Stoichiometry of the reaction:

5 moles of iron (II) ions (ferrous sulfate) reacts with 1 mole of permanganate ion (potassium permanganate)

So, 0.00762 moles of iron (II) ions (ferrous sulfate) will react with = \frac{1}{5}\times 0.00762=0.00152mol of permanganate ion (potassium permanganate)

Now, calculating the volume of permanganate ion (potassium permanganate) by using equation 1, we get:

Molarity of permanganate ion (potassium permanganate) = 0.152 M

Moles of permanganate ion (potassium permanganate) = 0.00152 mol

Putting values in equation 1, we get:

0.152mol/L=\frac{0.00152mol}{\text{Volume of permanganate ion (potassium permanganate)}}\\\\\text{Volume of permanganate ion (potassium permanganate)}=\frac{0.00152mol}{0.152mol/L}=0.01L=10.0mL

Hence, the volume of permanganate ion (potassium permanganate) is 10.0 mL

3 0
3 years ago
A student was comparing the solubility of equal amounts of table salt and table sugar at different temperatures.The table below
Mama L [17]
The sample that has the highest solubility is THE FIRST SAMPLE [TABLE SUGAR NO 1].
Solubility refers to the quantity of a solute that will dissolve in a given volume of solvent at a given temperature and pressure. The question above tells us that equal amounts of sugar and table salts were used. But looking at the table given in the question, you will see that sample 1 has the highest amount of solute that dissolve, that is 80,  the rest of the samples have values that are lower than that. 
8 0
3 years ago
Read 2 more answers
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