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
nata0808 [166]
3 years ago
7

What does it mean for an element to be oxidized?

Chemistry
1 answer:
Delicious77 [7]3 years ago
7 0
Oxgen has been added to the element
You might be interested in
The highest energy level of sodium contains ________ electron(s).
Effectus [21]

Answer:

A

Explanation:

The group number of an element is equal to the number of electrons the outermost shell (or highest energy level) contains

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
SOMEONE PLS HELP ME ITS URGENT
Simora [160]

If copper is heated with iron oxide there is no obvious reaction because

copper is less reactive than iron.

On a reactivity chart, copper is far below iron. This makes it impossible for a replacement reaction to occur, so the equation doesn't change.

I hope I helped!

5 0
3 years ago
Read 2 more answers
Which of the following equations has the correct products and is balanced correctly for a reaction between Na3PO4 and KOH?
ASHA 777 [7]

The answer is: A) Na3PO4 + 3KOH → 3NaOH + K3PO4, because K retains the same charge throughout the reaction.

This chemical reaction is double displacement reaction - cations (K⁺ and Na⁺) and anions (PO₄³⁻⁻ and OH⁻) of the two reactants switch places and form two new compounds.  

Na₃PO₄ is sodium phosphate.  

KOH is potassium hydroxide.  

NaOH is sodium hydroxide.  

K₃PO₄ is potassium phosphate.  

According to the mass conservation law, there are same number of atoms on both side of balanced chemical reaction.

5 0
3 years ago
The Goodyear blimp contains 5.7 x 10^6 L of helium at 25 degrees Celsius and 1 atm. What is the mass in grams of the helium insi
anygoal [31]

Answer:

1.72x10⁻⁵ g

Explanation:

To solve this problem we use the PV=nRT equation, where:

  • P = 1 atm
  • V = 5.7x10⁶ L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 25 °C ⇒ (25+273.16) = 298.16 K

And we <u>solve for n</u>:

  • 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
  • n = 4.29x10⁻⁶ mol

Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:

  • 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g

8 0
3 years ago
Other questions:
  • Why must the ph ammonia buffer solution be stored and dispensed in the fume exhaust hood?
    15·1 answer
  • What is the approximate pH of a .06 M solution of CH3COOH given that Ka= 1.78*10-5
    12·1 answer
  • What are the horizontal (from left to right) parts of the periodic table called?
    10·2 answers
  • Which societal problem can be solved in part by scientist identifying areas of high biodiversity
    14·2 answers
  • Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter freq
    14·1 answer
  • How to write the ionic formula for the K1+, MnO41- pair of ions
    5·1 answer
  • Can you always tell by looking at a substance if it is pure or not? ​
    10·1 answer
  • How much energy (joules) is needed to heat 250 grams of copper from 22 °C to 99 °C? The specific heat capacity (C) of copper is
    5·1 answer
  • What happen when aniline is treated with Chloroforan<br>presence of alcohol koh so<br>lution​
    5·1 answer
  • Calculate the volume in (L) of 1.13x10^5 grams of liquid sodium which has a density of 0.929g/cm3
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!