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Soloha48 [4]
2 years ago
5

Given the following information:

Chemistry
1 answer:
bonufazy [111]2 years ago
6 0

Answer:

Ya'll should really practice more because things like these are easy

Explanation:

You might be interested in
Carbon reacts with steam (h2o) at high temperatures to produce hydrogen and carbon monoxide. if 2 mol of carbon are exposed to 3
Ne4ueva [31]
The balanced equation for the above reaction is as follows;
C + H₂O ---> H₂ + CO
stoichiometry of C to H₂O is 1:1
1 mol of C reacts with 1 mol of H₂O
we need to find which is the limiting  reactant
2 mol of C and 3.1 mol of H₂O
therefore C is the limiting reactant and H₂O is in excess.
stoichiometry of C to H₂ is 1:1
then number of H₂ moles formed are equal to C moles reacted
number of H₂ moles formed = 2 mol
4 0
3 years ago
Chloroform is a liquid once used for anesthetic. What is the volume of 5.0 g of chloroform? The density of chloroform 1.49 g/mL
mario62 [17]

Answer:

3.3557047 mL

Explanation:

The density can be found using the following formula:

d=\frac{m}{v}

Let's rearrange the formula to find the volume, v.

d*v=\frac{m}{v}*v

d*v=m

\frac{d*v}{d} =\frac{m}{d}

v=\frac{m}{d}

The volume can be found by dividing the mass by the density. The mass of the chloroform is 5 grams and the density is 1.49 grams per milliliter. Therefore,

m= 5g \\d= 1.49 g/mL

Substitute the values into the formula.

v=\frac{5g}{1.49 g/mL}

Divide. When we divide, the grams, or g, in the numerator and denominator will cancel out.

v= \frac{5}{1.49}mL

v=3.3557047 mL

The volume of 5 grams of chloroform is 3.3557047 milliliters

4 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
2 years ago
Boron has primarily two isotopes, one with an atomic mass of 11.0 amu and another with an atomic mass of 10.0 amu. If the abunda
Masja [62]

Answer:

The atomic mass of the boron atom would be <em>10.135</em>

Explanation:

This is generally known as relative atomic mass.

Relative atomic mass or atomic weight is a physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to the atomic mass of 1/12 of the mass of a carbon-12 atom. Since both quantities in the ratio are masses, the resulting value is dimensionless; hence the value is said to be relative and does not have a unit.

<em>Note that the relative atomic mass of atoms is not always a whole number because of it being isotopic in nature.</em>

  • <em>Divide each abundance by 100 then multiply by atomic mass</em>
  • <em>Do that for each isotope, then add the two result. Thus</em>

Relative atomic mass of Boron = (18.5/100 x 11) + (81/100 x 10)

                                                 = 2.035 + 8.1

                                                 = 10.135

5 0
2 years ago
Properties of waves poem plzz help!!!!!!
GalinKa [24]

Answer:

They include frequency, period,speed,amplitude and phase

3 0
3 years ago
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