1. Write balanced equations for each of the following processes.

a. Chromium- 51, which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture.

b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a beta particle.

2. In each of the following nuclear reactions, supply the missing particle?
a. ^60Co -> ^60Ni + ?
b. ^97Tc + ? -> ^97Mo
c. ^99Tc -> ^99Ru + ?
d. ^239Pu -> ^235U + ?

3. A chemist wishing to do an experiment requiring 47Ca (half life = 4.5 days) needs 5.0ìg of the
nuclide. What mass of 47Ca must be ordered if it takes 48 h for delivery from the supplier? Assume that the atomic mass of ^47Ca iis 47.0

4. What is the ratio of ^209Pb/^238U by mass in a rock that is 4.5x10^9 years old? (For 238 U, t1/2= 4.5x10^9 years)

Please show me how you get the answers.

aerobic respiration

1.

a. Chromium-51 decays by electron capture, which means it captures an inner electron and combines it with a proton in the nucleus. The balanced equation for this process is:

^51Cr + e- -> ^51V + ν

b. Iodine-131 decays by producing a beta particle, which is an electron. The balanced equation for this process is:

^131I -> ^131Xe + β-

2.
a. The element Cobalt-60 decays to Nickel-60 by emitting a beta particle. So, the missing particle is a beta particle (β-).

b. Technetium-97 decays to Molybdenum-97. To balance the equation, we need to supply a neutron (n) to conserve both mass number and atomic number.

c. Technetium-99 decays to Ruthenium-99. To balance the equation, we need to supply an electron (e-) to maintain atomic number.

d. Plutonium-239 undergoes nuclear fission and splits into Uranium-235 and another particle. So, the missing particle is a neutron (n).

3. In order to determine the mass of 47Ca needed, we need to use the equation for radioactive decay:

m = m₀ * (1/2)^(t/t₁/₂)

Given that the half-life (t₁/₂) of 47Ca is 4.5 days and the delivery time (t) is 48 hours, we need to convert 48 hours to days:

t = 48 hours / 24 hours/day = 2 days

Plugging the values into the equation, we have:

m = 5.0 μg * (1/2)^(2/4.5)

Calculating this, we find:

m ≈ 3.11 μg

Therefore, approximately 3.11 μg of 47Ca must be ordered.

4. The ratio of ^209Pb/^238U can be calculated using the concept of radioactive decay and the decay constant:

R = (N0/N1) = (λ0/λ1) = (t1/2,0/t1/2,1) * (1 - exp(-λ1 * t)) / (1 - exp(-λ0 * t))

Given that the rock is 4.5 × 10^9 years old and the half-life of ^238U is 4.5 × 10^9 years, we can calculate the ratio:

R = (4.5 × 10^9 years / 4.5 × 10^9 years) * (1 - exp(-λ1 * 4.5 × 10^9 years)) / (1 - exp(-λ0 * 4.5 × 10^9 years))

Simplifying, we get:

R ≈ 1

Therefore, the ratio of ^209Pb/^238U by mass in the rock is approximately 1.

Frankly this is too many posts on one try AND it is far too much work. Here is the principle by which all of these work.

The subscripts must add up.
The superscripts must add up.
The element is determined by the atomic number. Example of what I mean.
I know the atomic number is supposed to go on the left hand side as a subscript and that can be done easily. The mass number is supposed to go on the left of the element symbol as a superscript which can't be done on these boards; therefore, I will write it as a superscript but on the right hand side of the symbol for the element.
26Cr51 + EC (electron capture)==> 23V51

Note that 23 subscript adds to the same on both sides, 51 as a superscript adds to the same on both sides. This is a different kind of decay but it works this way. An electron from the K shell drops into the nucleus. The Cr-51 nucleus has 24 protons(+ charge) and 27 neutrons (note 24 + 27 = 51.....I know Cr has 24 protons because the atomic number is 24 and I know it has 27 neutrons because 51(from the problem) - 24 = 27). So when the electron drops in, we now have 24+, 1-(from the electron) and 27 N(zero charge). One proton combines with the electron to make a neutron so we now have 23 protons and 28 neutrons. The symbol for 23 protons is V and it still has a mass number of 51 (23+28=51).

Another:
27Co60 ==> 28Ni60 + ?
Now you add up both sides. ? must have a subscript of -1 (because we have 27 on the left and 28 on the right and 28+(-1) = 27). What mass must it have. We have 60 on the left and 60 on the right; therefore, the mass number must be 0. What is it that has a charge of -1 and mass of zero? That must be an electron. So we would write ? as
-1eo

For the decay problems use
k = 0.693/t1/2 and
ln(No/N) = kt
That combination will work most decay problem involving half life and time.
Post your work if you get stuck. Or ask a particular question if there is something here you don't understand.