Which statement best summarizes this part of kafka's the metamorphosis?

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

3 years ago

On a potential energy diagram for the following processes, which of the following has an increase in entropy?

daser333 [38]

Hello,

I believe that your answer would be B. water freezes
Hope this helps

7 0

3 years ago

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Which state of matter would be described as having highly energized, charged particles that move extremely fast

uranmaximum [27]

There are four states of matter, solid, liquid, gas and plasma. Their formation is as when solid is heated it converts into liquid, liquid on heating converts into gases and gases on heating converts into plasma.

Plasma:
Plasma is the fourth state of matter. It is the highest energy state of matter.

Composition:
Plasma is made up of negatively charged and positively charged particles.

Result:
The answer to your question is Plasma.

8 0

3 years ago

ANSWER ASAP!

Rainbow [258]

Answer:

All around you there are chemical reactions taking place. Green plants are photosynthesising, car engines are relying on the reaction between petrol and air and your body is performing many complex reactions. In this chapter we will look at two common types of reactions that can occur in the world around you and in the chemistry laboratory. These two types of reactions are acid-base reactions and redox reactions.

Explanation:

7 0

3 years ago

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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