CHEMISTRY HIGH SCHOOL

An important application of exponential functions is working with half-life of radioactive isotopes in chemistry. These isotopes emit particles and decay into stable forms, in doing so they lose mass over time. Half-life of an isotope is the time it takes for the amount to decay by half. For example, the half life of Bromine-85 is 3 minutes. This means if you start with 60g of Br-85, 3 minutes later 30g will remain. How much Br-85 will remain after 20 minutes?

Answers

Answer 1

Answer:

Answer : The amount left after 20 minutes is, 0.592 grams.

Explanation :

Half-life of Bromine-85 = 3 min

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{3\text{ min}}$

$k=0.231\text{ min}^(-1)$

Now we have to calculate the amount left after decay.

Expression for rate law for first order kinetics is given by:

$t=(2.303)/(k)\log(a)/(a-x)$

where,

k = rate constant

t = time taken by sample = 20 min

a = initial amount of the reactant = 60 g

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$20=(2.303)/(0.231)\log(60)/(a-x)$

$a-x=0.592g$

Therefore, the amount left after 20 minutes is, 0.592 grams.

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A sample contains 10.5 g of the radioisotope Pb-212 and 157.5 g of its daughter isotope, Bi-212. How many half-lives have passed since the sample originally formed?4
14
15
147.5

Answers

Answer: The sample must have passed 4 half-lives after the sample was originally formed.

Explanation: This is a type of radioactive decay and all the radioactive process follow first order kinetics.

Equation for the reaction of decay of $_(82)^(212)\textrm{Pb}$ radioisotope follows:

$_(82)^(212)\textrm{Pb}\rightarrow _(83)^(212)\textrm{Bi}+_(-1)^0\beta$

To calculate the initial amount of $_(82)^(212)\textrm{Pb}$ , we will require the stoichiometry of the reaction and the moles of the reactant and product.

Expression for calculating the moles is given by:

$\text{no of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of $_(82)^(212)\textrm{Pb}$ left = $(10.5g)/(212g/mol)=0.0495moles$

Moles of $_(83)^(212)\textrm{Bi}=(157.5g)/(212g/mol)=0.7429moles$

By the stoichiometry of above reaction,

1 mole of $_(83)^(212)\textrm{Bi}$ is produced by 1 mole $_(82)^(212)\textrm{Pb}$

So, 0.7429 moles of $_(83)^(212)\textrm{Bi}$ will be produced by = $(1)/(1)* 0.7429=0.7429\text{ moles of }_(82)^(212)\textrm{Pb}$

Amount of $_(82)^(212)\textrm{Pb}$ decomposed will be = 0.7429 moles

Initial amount of $_(82)^(212)\textrm{Pb}$ will be = Amount decomposed + Amount left = (0.0495 + 0.7429)moles = 0.7924 moles

Now, to calculate the number of half lives, we use the formula:

$a=(a_o)/(2^n)$

where,

a = amount of reactant left after n-half lives = 0.0495 moles

$a_o$ = Initial amount of the reactant = 0.7924 moles

n = number of half lives

Putting values in above equation, we get:

$0.0495=(0.7924)/(2^n)$

$2^n=16.0080$

Taking log on both sides, we get

$n\log2=\log(16.0080)\nn=4$

Answer:

Explanation:

Edg 2020

What is water classified as

Answers

Water, or H2O, is a compound composed of the elements hydrogen (H) and oxygen (O).

Sulfur oxides Carbon monoxides Chlorofluorocarbons (CFCs)Nitrates are responsible for the hole in the ozone layer.

Answers

The best answer for the question above would be the chloroflourocarbons or the CFCs. These chloroflourocarbons or CFCs are the ones responsible for the depletion of the ozone - which leads to leaving a hole in its layer. These gases eat out the ozone layer and allows harmful UV rays of the sun to come in the Earth.

State and explain the trends in the atomic radius and ionization energy for elements Li to Cs.

Answers

Explanation:

ionisation energy decrease down the group as the atomic radius increases. Nuclear charge increases. Number of shell increases so the electron will experience more shielding so it would be easier for the atom to loss electron.

Final answer:

The atomic radius increases as you move down a group from Li to Cs, while the atomic radius generally decreases as you move across a period from Li to Cs. The ionization energy decreases down a group and increases across a period.

Explanation:

The atomic radius is the size of an atom, while ionization energy is the energy required to remove an electron from an atom. In general, as we move down a group from Li to Cs, the atomic radius increases due to the addition of more energy levels. This is because the electrons occupy higher energy orbitals farther away from the nucleus. On the other hand, as we move across a period from Li to Cs, the atomic radius generally decreases. This is because the effective nuclear charge increases, pulling the electrons closer to the nucleus.

Regarding ionization energy, it generally decreases down a group from Li to Cs. This is because the atomic radius increases, therefore making it easier to remove an electron from a larger, higher energy orbital. Conversely, as we move across a period, the ionization energy generally increases. This is because the atomic radius decreases, and the electrons are held more tightly by the increasing nuclear charge.

Learn more about Trends in atomic radius and ionization energy here:

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A car traveled at an average speed of 60 mph for two hours. How far did it travel? A.
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B.
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C.
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D.
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i need help and please answer

Answers

C-120 hope this helps

C.) Average speed = total distance/total time. Total time= avg speed x time = 60x2 = 120miles.

In Experiment 1, the density of pure ethanol was found to be:Question 1 options:less than 0.793 g/mL.0.793 g/mL.0.999 g/mL.greater than 0.999 g/mL.

Answers

It is 21 due to math a marica

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