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SVETLANKA909090 [29]
2 years ago
14

Which best defines partial pressure in a mixture of gases?

Chemistry
2 answers:
VladimirAG [237]2 years ago
7 0

Answer:

Pressure that is exerted by one gas as if it occupied a container by itself.

Oxana [17]2 years ago
4 0

Answer:

Pressure that is exerted by one gas as if it get occupied a container by itself

Explanation:

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A energia térmica corresponde à soma das energias cinéticas associadas a todas as partículas que compõem um corpo. Essa afirmati
Snezhnost [94]

answer

( )Errada is the answer

explanation

8 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 the electronegativity difference for a bond between potassium and iodine? A. 3.3 B. 4.6 C. 0.4 D. 1.7
user100 [1]
Iodine has an electronegativity of 2.5, and potassium has an electronegativity of 0.8, so the difference is:
2.5 - 0.8 = 1.7
7 0
3 years ago
Read 2 more answers
Part A: You are given a beaker containing a solution of an enzyme that hydrolyzes adenosine triphosphate (ATP) while it generate
Oksi-84 [34.3K]

Answer:

1) magnesium chloride

2) b) The copper is getting oxidized from Cu+ to Cu2+ and turns blue.

Explanation:

The work published by David N. Frick, Sukalyani Banik, and Ryan S. Rypma in J Mol Biol. 2007 Jan 26; 365(4): 1017–1032 clearly shows that divalent metal ions of group 2 such as Mg^2+ play an important role in ATP hydrolysis. Addition of EDTA decreased the rate of hydrolysis of ATP (due to sequestration of the divalent ion of group 2) indicating an active participation of divalent ions in the process.

2) The copper I ion is colourless because it is a d^10 specie. However, when it is oxidized to Cu^2+, a blue colour appears in the solution.

4 0
3 years ago
Volume of water displaced = 0.175 L Mass g Density g/cm3 Hardness: = Determine the unknown mineral. Unknown mineral: 7​
erastovalidia [21]

Answer:

Mass: 981.0 g

Density: 5.61 g/cm^3

Hardness: = 2.5 - 3

Unknown material: Chalcocite

Explanation:

5 0
3 years ago
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