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Sever21 [200]
3 years ago
9

When iridium-192 is used in cancer treatment, a small cylindrical piece of 192ir, 0.6 mm in diameter and 3.5 mm long, is surgica

lly inserted into the tumor. if the density of iridium is 22.42 g/cm3, how many iridium atoms are present in the sample?
Chemistry
1 answer:
Ivan3 years ago
4 0

Iridium-192 is used in cancer treatment, a small cylindrical piece of 192 Ir, 0.6 mm in diameter (0.3mm radius) and 3.5 mm long, is surgically inserted into the tumor. if the density of iridium is 22.42 g/cm3, how many iridium atoms are present in the sample?

Let us start by computing for the volume of the cylinder. V = π(r^2)*h where r and h are the radius and height of the cylinder, respectively. Let's convert all given dimensions to cm first. Radius = 0.03 cm, height is 0.35cm long.

V = π * (0.03cm)^2 * 0.35 cm = 9.896*10^-4 cm^3

Now we have the volume of 192-Ir, let's use the density provided to get it's mass, and once we have the mass let's use the molar mass to get the amount of moles. After getting the amount of moles, we use Avogadro's number to convert moles into number of atoms. See the calculation below and see if all units "cancel":

9.896*10^-4 cm^3 * (22.42 g/cm3) * (1 mole / 191.963 g) * (6.022x10^23 atoms /mole)

= 6.96 x 10^19 atoms of Ir-122 are present.

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For+the+reaction+H2+++I2+-+2HI+the+equilibrium+constant,+kc+is+49+at+a+fixed+temperature.+Two+mole+of+hydrogen+and+two+moles+of+
Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

    H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0
3 years ago
What is the density of a substance with a mass of 1.8 kg and a volume of 0.2 L?
horrorfan [7]

Answer:

<h3>The answer is 9.0 kg/L</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass = 1.8 kg

volume = 0.2 L

We have

density  =  \frac{1.8}{0.2}  \\

We have the final answer as

<h3>9.0 kg/L</h3>

Hope this helps you

5 0
3 years ago
What happens when Potassium Hydroxide reacts with Sulphuric Acid?​
Ede4ka [16]

As we know that neutralization reaction is a reaction in which base react with acid to form salt and water.

When Potassium Hydroxide reacts with Sulphuric Acid, it forms Potassium Sulphate and Water.

As a result of neutralization reaction, Potassium Sulphate and Water is formed.

2KOH + H2SO4 ----> K2SO4 + 2H2O

Here, K2SO4 is found in aqueous medium in neutralization reaction. It is a neutral salt.

5 0
3 years ago
Read 2 more answers
The Doppler effect related change in wave frequency to what?
Irina18 [472]

Answer:

Relative motion of source and observer.

Explanation:

Doppler Effect : It is the change in the observed frequency of the wave due to<em><u> relative motion of the source and object </u></em>.

8 0
3 years ago
If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0
3 years ago
Read 2 more answers
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