All categories

3 years ago

What is the half life of a radioactive isotope if 8g of a 32g sample remains unchanged after 12 hours?

Chemistry

1 answer:

Elodia [21]3 years ago

6 0

Answer:

You calculate the half-life from the amount of material that disappears in a given time.

Explanation:

Half-life (t½) is the time required for the nuclei to decay to half of the original amount.

Radioactive nuclei decay according to the equation:

Equation 1: Nt/N0=e^-λt, where

N0 is the initial number of nuclei at time t=0

Nt is the number of nuclei that remain after time t. We can also use any number that is proportional to the number of nuclei, such as mass or disintegration counts.

λ is a constant called the decay constant. Each nucleus has its own decay constant.

The equation for half-life is

Equation 2: t1/2=in2/λ

We can combine these two equations to get

Equation 3: Nt/N0=0.5^t/t½

EXAMPLE:

A 50 g sample of radium–226 decays to 5.7 g after 5000 years. What is the half-life of radium–226?

Solution:

Let’s use Equation 3:

Nt/N0=0.5^t/t½

5.7g/50g=0,5^5000yr/t1/2

0.114=0.5^5000yr/t1/2

Take the natural logarithm of each side

In 0.114= 5000yr/t1/2 × In 0.5

-2.7= 5000yr/t1/2 × (-0.693)

t1/2= (5000yr × 0.693/2.17) = 1600yr

The half-life of radium–226 is 1600 yr.

You might be interested in

The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38

adell [148]

Answer:

$1.21x10^{-3}$ M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x $10^{-3}$

H = $\frac{0.21}{1.38x10^{-3} }$

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = $1.21x10^{-3}$ M

7 0

4 years ago

The amount of energy available to do work after a chemical reaction has occurred is called

sveticcg [70]

Free energy is the answer i hope this helped

8 0

3 years ago

Calculate the volume of hydrogen if you have 12.1 moles of Hydrogen.

vodomira [7]

To Find :

The volume of 12.1 moles hydrogen at STP.

Solution :

We know at STP, 1 mole of gas any gas occupy a volume of 22.4 L.

Let, volume of 12.1 moles of hydrogen is x.

So, x = 22.4 × 12.1 L

x = 271.04 L

Therefore, the volume of hydrogen gas at STP is 271.04 L.

8 0

3 years ago

What mass of CH3COOH is present in a 250 mL cup of 1.25 mol/L solution of vinegar?

aleksley [76]

If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x

⇒ x = $\frac{250 ml * 1.25mol}{1000 ml}$

= 0.3125 mol

∴ moles of CH₃COOH in 250ml is 0.3125 mol

Now, Mass = mole × molar mass

= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol

= 18.75 g

∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>

4 0

3 years ago

What would be the properties of the 5th state of matter? (bosse-instein condensate)

kolbaska11 [484]

Answer:

a state of matter created when particles, called bosons, are cooled to near absolute zero (-273.15 degrees Celsius, or -460 degrees Fahrenheit).

Explanation:

hope this helps

pls mark brainliest

8 0

3 years ago