All categories

3 years ago

You have 40 g of a radioisotope. If the half-life of this radioisotope is 4 days, how many grams will remain after 12 days?

1 answer:

5 0

With reference to radioactive material, half-life is the time required to 50% depletion of initial amount of material.

Given: Initial amount of radioactive material = 40 g
Half life = 4 days.

Therefore, After 4 days, amount of compound left = 40/2 = 20 g
After 8 days, i.e 2 half-life, amount of compound left = 20/2 = 10 g
Finally after 12 days, i.e. 3 half-life, amount of compound left = 10/2 = 5 g

Thus, 5 <span>grams will remain after 12 days</span>

You might be interested in

Use your experimentally determined value of ksp and show,by calculations, that ag2cro4 should precipitate when 5ml of 0.004m agn

Doss [256]

When the value of Ksp = 3.83 x 10^-11 (should be given - missing in your Q)

So, according to the balanced equation of the reaction:

and by using ICE table:

Ag2CrO4(s) → 2Ag+ (Aq) + CrO4^2-(aq)

initial 0 0

change +2X +X

Equ 2X X

∴ Ksp = [Ag+]^2[CrO42-]

so by substitution:

∴ 3.83 x 10^-11 = (2X)^2* X

3.83 x 10^-11 = 4 X^3

∴X = 2.1 x 10^-4

∴[CrO42-] = X = 2.1 x 10^-4 M

[Ag+] = 2X = 2 * (2.1 x 10^-4)

= 4.2 x 10^-4 M

when we comparing with the actual concentration of [Ag+] and [CrO42-]

when moles Ag+ = molarity * volume

= 0.004 m * 0.005L

= 2 x 10^-5 moles
[Ag+] = moles / total volume

= 2 x 10^-5 / 0.01L

= 0.002 M

moles CrO42- = molarity * volume

= 0.0024 m * 0.005 L

= 1.2 x 10^-5 mol

∴[CrO42-] = moles / total volume

= (1.2 x 10^-5)mol / 0.01 L

= 0.0012 M

by comparing this values with the max concentration that is saturation in the solution

and when the 2 values of ions concentration are >>> than the max values o the concentrations that are will be saturated.

∴ the excess will precipitate out

8 0

3 years ago

You are working with a specific enzyme-catalyzed reaction in the lab. You are a very careful experimentalist, and as a result, a

lord [1]

B. It will increase the rate

Explanation:

In this experiment, the rate of the reaction will increase with increase in temperature. Since enzymes are chemical catalysts that speeds up the rate of chemical reactions, one must know that a high temperature favors the rate of chemical reaction.

At a high temperature, the catalyst action of the enzyme will increase rapidly.
Caution must be taken because at extremely high temperatures, the enzymes can become denatured.

Learn more:

proteins as enzymes: brainly.com/question/13022851

#learnwithBrainly

7 0

3 years ago

2] Calculate the percentage of the element forming sugar glucose C6H1206<br>[C=12, H=1, 0=16]

inysia [295]

Answer:

wouldn't it be 52

Explanation:

3 0

3 years ago

What is the concentration of a solution in which 10.0 g of AgNO3 is dissolved in 500 mL of solution?

Zigmanuir [339]

molar concentration of AgNO₃ solution = 0.118 mole/L

Explanation:

Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.

number of moles = mass / molecular weight

number of moles of AgNO₃ = 10 / 170 = 0.0588

molar concentration = number of moles / volume (L)

molar concentration of AgNO₃ solution = 0.0588 / 0.5

molar concentration of AgNO₃ solution = 0.118 mole/L

Learn more about:

molar concentration

brainly.com/question/1286583

#learnwithBrainly

6 0

4 years ago

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

6 0

3 years ago