What branch of Earth science studies the position of Earth in the solar

Consider the following reversible reaction. 2H2O(g)<—>2H2(g)+O2(g) What is the equilibrium constant expression for the giv

masha68 [24]

The reaction: 2H2(g) + O2(g) → 2H2O(g), can be interpreted as: a. 2 moles of hydrogen gas reacts with 1 mole of oxygen gas to produce 2 moles of water.

7 0

3 years ago

Read 2 more answers

Which element does this Bohr model represent? (look at a periodic table if needed)

Serjik [45]

Circulating round the nucleus are the electrons in various orbits of different energy levels. Electrons are negatively charged and represented by the symbol 'e'. In the given image the number of protons are -6. Hence the element in question is Carbon as Carbon has the atomic number 6.

6 0

2 years ago

the law of conversation of (mass/energy) states that the total mass before a chemical reaction is the same as the totally mass a

Nimfa-mama [501]

The law of conservation of mass says that matter cannot be created or destroyed by ordinary chemical or physical changes.

This means:

The mass will be the same before and after the chemical or physical change

There are the same number of each type of atom before and after the change

The total mass of all components of a chemical reaction or physical change can be measured before and after the change to demonstrate that the mass is constant.

Scientists believe that energy is conserved, like mass, during a physical or chemical change. The law of conservation of energy says that energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical changes.

Energy can be more challenging for scientists to keep track of during a chemical reaction than mass. Mass is able to be measured on a balance before and after the change, but scientists must use more advanced techniques to measure the energy in a system.

Energy is converted from one form to another during a chemical change.This energy is not created or destroyed; it is just converted from one form of energy to another during the chemical and physical changes that occur.

Physical Change: In a physical change, atoms are rearranged but remain in the same sample of matter. The same numbers of atoms are present before and after the change, and they have the same total mass. Atoms are spread farther apart or pushed closer together during a phase change. When a substance changes phases, the volume may change but the mass doesn't.

Chemical Change: In a chemical change, the original bonds are broken and new bonds are formed. The same atoms are present before and after a chemical change, they are just rearranged to make different compounds.

5 0

3 years ago

A

amm1812

Solution:

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

Fe2+ ---> Fe3+
S2¯ ---> SO42¯
NO3¯ ---> NO

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.

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

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:

3 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]

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:

Sb26+ + CO32- + C ---> Sb + CO

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:

Sb26+ ---> Sb
CO32- ---> CO
C ---> CO

3) Balance as if in acidic solution:

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

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

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!

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

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

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

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

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:

Cr3+ ---> CrO42¯
I33¯ ---> IO4¯
H2O2 ---> H2O

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:

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

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:

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

4) Equalize the electrons:

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

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

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

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:

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

3 0

3 years ago

What products result from mixing aqueous solutions of Cr(NO3)2(aq) and NaOH(aq)? Question 10 options: Cr(OH)2(s), Na+(aq), and N

Alecsey [184]

Answer:

Cr(OH)2(s), Na+(aq), and NO3−(aq)

Explanation:

Let is consider the molecular equation;

2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)

This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.

Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.

4 0

3 years ago