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3 years ago

if the density of a certain spherical atomic nucleus is 1.0x10^14 g cm^-3 and its mass is 2.0x10^-23 g, what is it radius in cm?

2 answers:

6 0

Density as you should know is equal to mass/volume. Solve for the volume.

Now, you know that you are dealing with a sphere. You know the volume of a sphere is V= 4/3pi r ^3.

You know the volume so just solve for r

lesantik [10]3 years ago

4 0

Answer:

r = $3.63x10^{-13}$ cm

Explanation:

The density (d) is the mass divided by the volume, so:

d = m/V

$1.0x10^{14} = \frac{2.0x10^{-23}}{V}$

$V = \frac{2.0x10^{-23}}{1.0x10^{14}}$

V = $2.0x10^{-37} cm^3$

The volume of a sphere is

$V = \frac{4xpixr^3}{3}$

For pi = 3.14

$2.0x10^{-37} = \frac{4x3.14xr^3}{3}$

$12.56r^3 = 6.0x10^{-37}$

$r^3 = 4.78x10^{-38}$

$r = \sqrt[3]{4.78x10^{-38}}$

r = $3.63x10^{-13}$ cm

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svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ, ΔSºrxn = 0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG = 1.8 x 10⁴ J

K = 8.2 x 10⁻²

Explanation:

a) C6H5−CH2CH3 ⇒ C6H5−CH=CH2 + H₂

ΔHf kJ/mol -12.5 103.8 0

ΔGºf kJ/K 119.7 202.5 0

Sº J/mol 255 238 130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate ΔHºrxn we have

ΔHºrxn = ΔHfº C6H5−CH=CH2 + ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn = 103.8 kJ + 0 kJ - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 + ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn= 202.5 kJ + 0 kJ - 119.7 kJ = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn = ΔHrxn -TΔS

we see that will happen when the term TΔS becomes greater than ΔHrxn since ΔS is positive , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that ΔºHrxn and ΔSºrxn remain constant at the higher temperature and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K = (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

ΔGrxn = ΔHrxn -TΔS

to calculate ΔGrxn with the assumption that ΔHº and ΔSºremain constant.

ΔG = 116.3 kJ - (600+273 K) x 0.113 kJ/K = 116.3 kJ - 873 K x 0.113 kJ/K

ΔG = 116.3 kJ - 98.6 kJ = 17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K x 873 K)) = 8.2 x 10⁻²

8 0

2 years ago

How does a sample of hydrogen at 10 °C compare to a sample of hydrogen at 350 K?

Aleks04 [339]

Answer: -

The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

Explanation: -

Temperature of the hydrogen gas first sample = 10 °C.

Temperature in kelvin scale of the first sample = 10 + 273 = 283 K

For the second sample, the temperature is 350 K.

Thus we see the second sample of the hydrogen gas more temperature than the first sample.

We know from the kinetic theory of gases that

The kinetic energy of gas molecules increases with the increase in temperature of the gas. The speed of the movement of gas molecules also increase with the increase in kinetic energy.

So higher the temperature of a gas, more is the kinetic energy and more is the movement speed of the gas molecules.

Thus the hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

8 0

2 years ago

Read 2 more answers

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

2 years ago