The molar nuclear mass of carbon-14 is 14.003241 g/mol. The molar mass of a prot

Question

The molar nuclear mass of carbon-14 is 14.003241 g/mol. The molar mass of a proton is 1.007825 g/mol. The molar mass of a neutron is 1.008665 g/mol. Calculate the binding energy of C-14. (c = 3.00 ´ 108m/s)

Which symbol represents a positron?

The following reaction occurs spontaneously.

3 Cu2+(aq) + 2 Fe(s) ® 2 Fe3+(aq) + 3 Cu(s)

Write the balanced oxidation half-reaction.

The following reaction occurs spontaneously.

2 H+(aq) + Ca(s) ® Ca2+(aq) + H2(g)

Write the balanced reduction half-reaction.

The half-life of iodine-123 is 13.2 hours. What percentage of iodine-123 remains in a sample after 36 hours?

By what (single step) process does radon-226 decay to francium-236?

Consider the following half-reactions:

F2(g) + 2 e- ® 2 F-(aq) E° = +2.87 V
Cu2+(aq) + 2 e- ® Cu(s) E° = +0.34 V
Sn2+(aq) + 2 e- ® Sn(s) E° = -0.14 V
Al3+(aq) + 3 e- ® Al(s) E° = -1.66 V
Na+(aq) + e- ® Na(s) E° = -2.71 V

Which of the above elements or ions will oxidize Sn(s)?

What is the correct cell notation for a voltaic cell based on the reaction below?

Ni2+(aq) + Zn(s) ® Ni(s) + Zn2+(aq)

Consider the following half-reactions:

Cu2+(aq) + 2 e- ® Cu(s) E° = +0.34 V
Sn2+(aq) + 2 e- ® Sn(s) E° = -0.14 V
Fe2+(aq) + 2 e- ® Fe(s) E° = -0.44 V
Al3+(aq) + 3 e- ® Al(s) E° = -1.66 V
Mg2+(aq) + 2 e- ® Mg(s) E° = -2.37 V

Which of the above metals or metal ions will reduce Fe2+(aq)?

Assuming the following reaction proceeds in the forward direction,

Ni2+(aq) + Cr(s) ® Ni(s) + Cr2+(aq)

Write a balanced chemical equation for the following reaction in a basic solution.

ClO3-(aq) + Mn(OH)2(s) ® Cl-(aq) + MnO2(s)

Uranium-235 has a half-life of 7.04 ´ 108 years. How many years will it take for 25% of a U-235 sample to decay?

1.

The molar nuclear mass of carbon-14 is 14.003241 g/mol. The molar mass of a proton is 1.007825 g/mol. The molar mass of a neutron is 1.008665 g/mol. Calculate the binding energy of C-14. (c = 3.00 ´ 108m/s)

(Points : 4)        1.13 ´ 10-1 J/mol
       1.13 ´ 10-4 J/mol
       1.02 ´ 1013 J/mol
       1.00 ´ 1016 J/mol
       1.00 ´ 1016 J/mol

Explanation / Answer

Protons mass in C-14 = 6 * 1.007825 = 6.04695
Neutrons mass in C-14 = 8 * 1.008665 = 8.06932
C-14 mass = 14.003241
dM = (6.04695+8.06932) - 14.003241
dm = 0.113029
E = dM * C^2
E = 0.113029*(2.9978*10^8)^2
E 1.01577*10^16 J

Symbol of positron is " +1 e 0 "

Tl+ + 2 Cl- ------> 2 Tl + Cl2
Here Tl+ undergone reduction so it acts as a cathode and Cl_ under gone oxidation so it acts as anode.
Cl2(g) + 2 e- ----> 2 Cl-(aq) ; E° = +1.36 V
Tl+(aq) + e- ----> Tl(s) ; E° = -0.34 V
E0 cell = -0.34-1.36
E0 cell = -1.7 V

3 Cu2+(aq) + 2 Fe(s) ® 2 Fe3+(aq) + 3 Cu(s)
3 Cu+2 -----> 3 Cu Reduction reaction reduction done by 6 e-
2 Fe -----> 2 Fe+3 oxidation reaction and it is done by 6 e-
so the half cell reaction = 2 Fe(s) -----> 2 Fe+3(aq) + 6e-

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