All categories

2 years ago

Phosphorus-33 has a half-life of 25 days. What fraction of the original sample would remain after 300 days?

Chemistry

1 answer:

zloy xaker [14]2 years ago

6 0

Answer:

The correct answer is - 1/4096.

Explanation:

In the starting, there is that the number of atoms is N 0 .

The number of half-lives in this case, = 300/ 25 = 12

After the first half-lives or 25 days, N 1 = N 0 /2

the half life is that time where half of the sample had decay.

After the second half-life or 50 days, N 2 = N 1 /2 = N 0 /4 .

There is division by two the original amount, then after four times you divide four times for 2 that means that you divide by 2 ^12 =4096 .

So the final amount remain is N 12 = N 0 /4096 or 1/4096.

You might be interested in

Calculate the concentration of the following solution in mol/dm3 0.1 moles of NaCl in 200 cm3

taurus [48]

1 $cm ^{3}$ = 0.001 $dm ^{3}$ . Therefore 200 $cm ^{3}$ = 0.2 $dm ^{3}$ . Molarity = $\frac{number of moles of NaCl}{volume of the solution} = \frac{0.1}{0.2} = 0.5 mol/dm^{3}$ . Hope this helps.

8 0

3 years ago

Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °

Mila [183]

This problem is to use the Claussius-Clapeyron Equation, which is:

ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]

Where p2 and p1 and vapor pressure at estates 2 and 1

ΔH is the enthalpy of vaporization

R is the universal constant of gases = 8.314 J / mol*K

T2 and T1 are the temperatures at the estates 2 and 1.

The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa

Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol

=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]

=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179

=> x = 314.6 K => 314.6 - 273.15 = 41.5°C

Answer: 41.5 °C

3 0

4 years ago

What two structures would provide a positive identification of a plant cell under a microscope? a)cell wall, mitochondria b)plas

olga55 [171]

Cell wall, Large central vacuole, Chloroplasts

3 0

3 years ago

Read 2 more answers

slader An experiment is carried out where 13.9 g of solid NaOH is dissolved in 250.0 g of water in a coffee-cup calorimeter. Dis

kvv77 [185]

Answer:

The mass of the surrounding is $M_t = 263.9 \ g$

Explanation:

From the question we are told that

The mass of $NaOH$ is $m = 13.9 \ g$

The mass of water is $m_w = 250.0g$

The chemical equation for the dissociation process is

$NaOH _{(s) } ---> Na^{+}_{(aq)} + OH^{-} _{(aq)}$

The specific heat capacity of the mixture is $c_p = 4.18 J g^{-1} C^{-1}$

The combined mass of the solution is

$M_t = 13 + 250$

$M_t = 263.9 \ g$

The mass of the surround here is the mass of the coffee-cup calorimeter and this contain the mixture ( water and the NaOH ) so the mass of the surrounding is

$M_t = 263.9 \ g$

3 0

3 years ago

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

nikdorinn [45]

Answer : The volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.