CHEMISTRY COLLEGE

Iridium-192 is used in medicine to treat prostate cancer. Iridium-192 has two modes of radioactive decay: 96% of the time it decays by beta emission and 4% of the time it decays by electron capture. What are the daughter nuclides of these two decay processes?

Answers

Answer 1

Answer:

Answer:

The daughter nuclides of these two decay processes are $^(192)_(78)Pt$ and $^(192)_(76)Os$ .

Explanation:

The beta emission is represented by:

A = (Z + 1) + (n - 1) = is invariant

n: neutron

p: proton

$^(A)_(Z)X \rightarrow ^(A)_(Z+1)Y$

Hence, the daughter nuclide of the beta emission of Ir-192 is:

$^(192)_(77)Ir \rightarrow ^(192)_(78)Pt$

Now, electron capture is represented by:

A = (Z - 1) + (n + 1) = is invariant

$^(A)_(Z)X \rightarrow ^(A)_(Z-1)Y$

Then, the daughter nuclide of the electron capture of Ir-192 is:

$^(192)_(77)Ir \rightarrow ^(192)_(76)Os$

Therefore, the daughter nuclides of these two decay processes are $^(192)_(78)Pt$ and $^(192)_(76)Os$ .

I hope it helps you!

Related Questions

A laboratory scientist wants to analyze the chlorogenic acid content of a 0.5 L sample. He uses an indicator solution and titrates with sodium hydroxide. He finds the equivalence point occurs when he's titrated 9.5 mL of 0.05 M sodium hydroxide. Approximately how much chlorogenic acid is in the sample? 475 umol ® 480 umol © 500 umol 510 umol
Calculate ΔHrxn for the following reaction: C(s) + H2O(g) --> CO(g) + H2(g) Use the following reactions and given ΔH values: C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ 2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ 2 H2 (g) + O2 (g) → 2 H2O (g), ΔH= -483.6 kJ
What would be the effect on the observed melting point of sample were poorly packed?
Select ALL factors in conservation: *Social conditions such as need for electricity, famine, and war.Scientific data related to the ecosystem and the effect of environmental changes.Political action by governments and other organizations such as environmentalprotection groups.Economic issues such as cost of wood products, fuel for heat, price of electricity, andincome levels of local people.
Select all the true statements. Group of answer choices The +3 oxidation state is characteristic of the actinides. All actinides are radioactive. Cerium (Ce) rnakes 100th in abundance (by mass %). Valence-state electronegativity is when a metal with a positive oxidation state has a greater attraction for the bonded electrons (thus a higher electronegativity) than it does when it has a 0 oxidation state. The actinides are silvery and chemically reactive. The lanthanides are in Period 7.

LAST ATTEMPT IM MARKING AS BRAINLIEST!! (What is this compound ? )

Answers

Answer:

its water h²O

Explanation:

Hopeitshelp

Unknown A melts at 113- 114oC. Known compounds 3-Nitroaniline and 4-Nitrophenol both melt at 112-114 oC. If A is mixed with 3-Nitroaniline and the melting point becomes broad and depressed, what must A be __________A) 3-Nitroaniline B) 4-Nitrophenol C) Both

Answers

Answer:

C) Both

Explanation:

Whenever we mix any pure form of a compound with some other form of a compound which is not in the other standard pure state, this results in the melting point of mixture to get dispersed and it becomes broad form.

Thus, when a known compound of 3-Nitroaniline mixes with both 3-Nitroaniline and 4-Nitrophenol, the melting point of the compound becomes depressed and board.

Thus the correct option is (C).

What are some examples of matter?

Answers

Answer:

An apple.

A person.

A table.

Air.

Water.

A computer.

Paper.

Iron.

Hope this helps you

Answer:

your boddy is made of mater and a clock too it is still a mater of time.

Explanation:

100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of volume 10.0 L at temperature 1508 K. The container is placed in an evacuated chamber and a small hole of area 3.23 mm2 is made in the upper container wall. After 2.00 hours, the mass of copper in the container has decreased by 1 0.0168 g. Assuming the mass loss is due to effusion, calculate the vapor pressure of liquid copper at 1508 K. Hint: because the liquid constantly evaporates, the pressure inside the container is constant

Answers

Answer:

8.912x10^-18

Explanation:

-dn/dt = pANa/2piMRT

100 g = initial copper

Number of moles = 100/63.546

= 1.5736

Mass of copper left = 100-10.0168

= 89.9832

Moles = 89.9832/63.546

= 1.4160

dn = 1.4160-1.5736

= -0.1576

dt = 2 hrs

A = 3.23mm² = 3.23x10^-6

M = 63.546

T = 0.0821

T = 1508k

Na = 6.023x10²³

When we insert all these into the formula above

We get

P = 8.912x10^-18atm

A piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm. (b) The gas is allowed to expand reversibly and isothermally to the same final volume. Please calculate the work done by the gas system in these two processes, respectively. Which process does more work? (revised from 6/e exercise 8.11) Please show calculation details.

Answers

Answer:

The second experiment (reversible path) does more work

Explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

(b) The gas is allowed to expand reversibly and isothermally to the same final volume.

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work

How many grams of sucrose must be added to 375 mL of watertoprepare a 2.75m/m percent solution of sucrose?

Answers

Answer : The mass of sucrose added to 375 mL of water must be, 10.6 grams.

Explanation :

As we are given that 2.75 m/m percent solution of sucrose. That means, 2.75 grams of sucrose present in 100 grams of solution.

Mass of solution = 100 g

Mass of sucrose = 2.75 g

Mass of water = Mass of solution - Mass of sucrose

Mass of water = 100 g - 2.75 g

Mass of water = 97.25 g

First we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}* \text{Volume of water}$

Density of water = 1.00 g/mL

Volume of water = 375 mL

$\text{Mass of water}=1.00g/mL* 375mL=375g$

Now we have to calculate the mass of sucrose in 375 g of water.

As, 97.25 grams of water contain 2.75 grams of sucrose

So, 375 grams of water contain $(375)/(97.25)* 2.75=10.6$ grams of sucrose

Therefore, the mass of sucrose added to 375 mL of water must be, 10.6 grams.

To make a 2.75% m/m sucrose solution, you need to add approximately 1062 grams of sucrose to 375 mL of water, considering the density of water as 1 g/mL.

To prepare a mass/mass (m/m) percent solution of sucrose, you need to calculate the mass of sucrose (in grams) that needs to be added to 375 mL of water to achieve a 2.75% concentration.

Here's how you can calculate it:

1. Convert the volume of water to grams, considering the density of water:

Density of water ≈ 1 g/mL

Mass of water = Volume of water × Density of water

Mass of water = 375 mL × 1 g/mL = 375 g

2. Determine the desired mass of sucrose as a percentage of the total mass:

Desired m/m percent = 2.75%

3. Calculate the mass of sucrose needed:

Mass of sucrose = (Desired m/m percent / 100) × Total mass

Mass of sucrose = (2.75 / 100) × (375 g + Mass of sucrose)

4. Rearrange the equation to solve for the mass of sucrose:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - (2.75 / 100))