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DIA [1.3K]
3 years ago
6

Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04

days. How much time is required for the activity of a sample of iodine-131 to fall to 17.5 percent of its original value?
Chemistry
1 answer:
gregori [183]3 years ago
5 0

Answer:

24.12 days

Explanation:

First off, let's find out how many decays the compound ; iodine131 undergoes to get to 17.5 percent of its original value.

Half life is simply the time required for a quantity to reduce to half of its initial value.

The half-life of iodine-131 is 8.04 days.

100% - 50%  (First Half life)

50%  - 25%  (Second Half life)

25%  - 17.5%  (Third Half life)

This mwans i would take three halff lives;

Time requred = 3 * Half life = 3 * 8.04 days = 24.12 days

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A gas occupies a volume of 2.4 L at 14.1 kPa. What volume will the gas occupy at 84.6 kPa?
Naddik [55]

Answer:

  • <u>0.40 L</u>

Explanation:

Boyle's law for gases states that, at constant temperature, the volume and pressure of a fixed amount of gas are inversely related.

Mathematically, that is:

  • PV = constant

  • P₁V₁ = P₂V₂

Here, you have:

  • V₁ = 2.4 L
  • P₁ = 14.1 Kpa
  • P₂ = 84.6 KPa
  • V₂ = ?

Then, you can solve for V₂:

  • V₂ = P₁V₁ / P₂

Substitute and compute:

  • V₂ = 14.1 KPa × 2.4L / 84.6 KPa = 0.40 L ← answer
3 0
3 years ago
What is the total number of Joules of heat absorbed by 65.0 grams of water when the temperature of the water is raised from 25°C
Norma-Jean [14]

Answer:

Total number of heat absorbed is 4.08kJ

Explanation:

Explanation is contained in the picture attached

5 0
3 years ago
NItrogen in air reacts at high temperature to form NO2 according to the reaction:
Rina8888 [55]

Answer:

The correct answer is option E.

Explanation:

Structures for the reactants and products are given in an aimage ;

Number of double bonds in oxygen gas molecule = 1

Number of double bonds in nitro dioxide gas molecule = 1

Number of single bond in in nitro dioxide gas molecule = 1

Number of triple bonds in nitrogen gas molecule = 1

N_2+2O_2\rightarrow 2NO_2,\Delta H=?

\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]

\Delta H_{f,NO_2}=33.18 kJ/mol

\Delta H_{f,N_2}=0 (pure element)

\Delta H_{f,O_2}=0 (pure element )

\Delta H=2 mol\times 33.18 kJ/mol=66.36kJ=15.86 kcal

The enthalpy of the given reaction is 15.86 kcal.

6 0
3 years ago
You operate a nuclear reactors and want to use fissionable mass that will sustain a slow control what type of mass will you use
puteri [66]

Answer: I believe it is critical mass

Explanation:

3 0
2 years ago
Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How ma
REY [17]

Answer:

34.7mL

Explanation:

First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.

So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn

All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.

So now we need to go to HCl!

To do that we multiply by the molar coefficients in the chemical equation:

\frac{0.0956g Zn}{1 } (\frac{2 mol HCl}{1molZn})

This leaves us with 2(0.0956) = 0.1912 mol HCl

Now we use the relationship M= moles / volume , to calculate our volume

Rearranging we get that V = moles / M

Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl

V= 0.0347 L

To change this to milliliters we multiply by 1000 so:

34.7 mL

7 0
4 years ago
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