##### Balance nuclear equations

 Chemistry Tutor: None Selected Time limit: 1 Day

Apr 9th, 2015

9038Sr → 0–1β + 9039Y                                   Use the conservation of mass and charge laws.

3819K → 01β + 3818Ar

21089Ac → 20687Fr + 42He

Apr 9th, 2015

Thank you! can you help me with another problem like this?

Apr 9th, 2015

Sure.

Apr 9th, 2015

Apr 9th, 2015

heres the question

Apr 9th, 2015

247100Fm → 42α + 24398Cf

3925Mn → 01e + 3924Cr

12650Sn → 12651Sb + 0–1e

Apr 9th, 2015

It says the second one is wrong

Apr 9th, 2015

the second one is suppose to be 49 on the Mn not a 39

Apr 9th, 2015

Sorry, it did not notice. Then

4925Mn → 01e + 4924Cr

Apr 9th, 2015

No worries! and thank you!

For this one:

Iodine-131 is a radioactive isotope. After 7.00 days, 54.7% of a sample of 131^I remains. What is the half-life of 131^I?

How would I do this? im not sure how.

Apr 9th, 2015

This is a different problem. The mass of a radioactive element decreases by an exponential law

m(t) = m_0 (1/2)^(t/T), where t time elapsed and T is the half-life.

Given m(7) = 0.547 m_0 = m_0 (1/2)^(7/T), we get (1/2)^(7/T) = 0.547.

Take the logarithm (log with the base 10, for example): (7/T) log(0.5) = log0.547

7/T = log(0.547)/log(0.5) or T = 7 log(0.5)/log(0.547) = 8.04 days.

Apr 9th, 2015

oh okay so I would have to use that equation to answer this kind of problem?

Apr 9th, 2015

For this question would I do the same thing? How would I do this? I tried answering this one but I keep getting it wrong.

Gold-198 has a half-life of 2.7 days. How much of a 498.3 mg gold-198 sample will remain after 13.5 days?

Apr 9th, 2015

The formula m(t) = m_0 (1/2)^(t/T) is the same, however, T is now known.

So, m(13.5) = 498.3 (1/2)^(13.5/2.7) = 498.3 / 2^5 = = 498.3/32 = 15.57 mg.

Apr 9th, 2015

Oh okay. Thank you! I found out what I did wrong. For this question would I use the equation with ln? For the 18 hour one I got 97.16% The 1 day I got 46.68%. The 5 half-lives I got 5.61%. But I got it wrong. Im not sure how to do it

Technetium-99m, an important radionuclide in medicine, has a half-life of 6.0 hours. What percentage of a sample of technetium-99m will remain undecayed after the following times?

18.0 hours?

1 day?

5 half-lives?

Apr 9th, 2015

Take the same formula m(t) = m_0 (1/2)^(t/T). Every half-life period the amount of the material will be halved.

If t = 18.0 hr, then m(18.0) = m_0 (1/2)^(18.0/6.0) = m_0 (1/2)^3 = (1/8) m_0

1/8 or 12.5% will be left

If t = 24 hr, then m(24) = m_0 (1/2)^(24/6) = (1/16) m_0

1/16 or 6.25% will be left

If 5 half-lifes (30 hr), then m(30) = m_0 (1/2)^5 = (1/32) m_0

1/32 or 3.125% will be left

Apr 9th, 2015

Thank you! I see what I did wrong. For this problem I put in 4 neutrons but its says I got it incorrect. Im not sure why I got it wrong

How many neutrons are needed to initiate the fission reaction shown?

(235/92)U+?(1/0)n ------> (137/52)Te+(97/40)Zr+2(1/0)n

Apr 9th, 2015

One will be enough (235 + 1 = 137 + 97 + 2).

Apr 9th, 2015

So it would be 236 neutrons? I put that in and it was wrong

Apr 9th, 2015

No, just 1 neutron needs to be added to a nucleus of uranium-235.

Apr 9th, 2015

okay thank you! sorry I keep asking questions. Im confused on this chapter

Apr 9th, 2015

For this one im not sure what to do. Do I have to add the atomic number and mass to get the value?

Balance this nuclear reaction by supplying the missing nucleus.

(249/98)Cf+(?/?)? ------> (263/106)Sg+4(1/0)n

Apr 9th, 2015

unknown mass x, unknown charge y

249 + x = 263 + 4; x = 18

98 + y = 106; y = 8 (oxygen)

Apr 9th, 2015

okay. so i would write it like (18/8)O

Apr 9th, 2015

yes, you are right

Apr 9th, 2015

okay. thank you!

Apr 9th, 2015

you are welcome

Apr 9th, 2015

Hi! Sorry Im asking here. My thing wont let me ask a new question. but I need help with reduction and oxidation reactions. Thank you.

Classify the following half-reactions as reduction half-reactions or oxidation half-reactions.

H2(g) ------> 2H^+ (aq) +2e^-

(1/2)O2(g)+2H^+(aq)+2e^- ------>H2O(g)

Cd(s)+2OH^-(aq) ----->Cd(OH)2(s)+2e^-

2NiO(OH)(s) +2H2O(l)+2e^- ------> 2Ni(OH)2(s)+2OH^- (aq)

Fe(s) ------> Fe^2+(aq)+2e^-

Apr 13th, 2015

1. Oxydation - loss of electrons

2. Reduction - gain of electrons

3. Oxydation

4. Reduction

5. Oxydation

Apr 13th, 2015

okay I see, thank you! for redox and non-redox question would I do the same thing?

Classify each of the following as a redox reaction or a non-redox reaction.

SO3+H2O ------>H2SO4

Zn+CuCl2 ---->ZnCl2+Cu

2CO+O2 ------>2CO2

HCl+NaOH ----->H2O+NaCl

Apr 13th, 2015

Never mind, I figured it out. Thank you!

Apr 13th, 2015

You are welcome.

Apr 13th, 2015

Hi again! I need help with some chemistry questions. Sorry Im asking here again, something is wrong with my account and wont let me ask a new question.

Calculate the E°cell for the following equation.

Cu(s)+Ag^+(aq)----->Cu^+(aq) +Ag(s)

Apr 16th, 2015

im not sure how to do this one

Apr 16th, 2015

There are two half-reactions: Cu(s) -> Cu^+ (aq) + e^- (oxydation, anode, reduction potential E anode = 0.52 V) and Ag^+ (aq) + e^- -> Ag (s) (reduction, cathode, reduction potential E cathode = 0.80 V. The cell will give E cathode - E anode = 0.80 - 0.52 = 0.28 V for a standard cell emf.

Apr 16th, 2015

Okay! I would do the same thing for this? I got 0.35 but not sure if its right

Sn(s)+F2(g) -----> Sn^2+(aq)+2F^-(aq)

Apr 16th, 2015

F2(g) + 2e^- -> 2F^- (reduction, cathode, Ec= +2.87 V)

Sn (s) -> Sn^2+ (aq) + 2e^- (oxidation, anode, Ea= 0.14 V)

E^0 cell = 2.87 - 0.14 = 2.73 V

Apr 16th, 2015

Its says its incorrect

Apr 16th, 2015

"Incorrect. You may have summed two reduction potentials. You need to flip the sign of the potential for the oxidation half-reaction before adding it to the potential for the reduction half-reaction."

This is what it said when i got it wrong

Apr 16th, 2015

Never mind I figured it out!

Apr 16th, 2015

For the following reactions, determine E°, ΔG°, and K, given the balanced half reactions and the standard reduction potentials. Also, determine if the reaction is spontaneous as written.

a.) 2Co^3+ + H3AsO3+H2O ------> 2Co^2+ +H3AsO4+2H^+

Co^3+ +e^- ----->Co^2+                                       E°=1.92 V

H3AsO4+2H^+ +2e^- ----->H3AsO3+H2O           E°=0.575 V

b.) 4Fe^3+ +2H2O -----> 4Fe^2+ +O2+4H^+

Fe^3+ +e^- -----> Fe^2+                            E°=0.771 V

(1/2)O2+2H^+ +2e^- -------> H2O                E°=1.229 V

I keep getting this wrong and I have no idea what I am doing wrong.

Apr 16th, 2015

I still cant figure it out...

Apr 16th, 2015

For the question above I seem to keep getting it incorrect again. I tried to look at some help videos but still could not understand it.

Apr 17th, 2015

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Apr 9th, 2015
May 23rd, 2017
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