The radioactive decay obeys first order kinetics
the rate law expression for radioactive decay is
![ln\frac{[A_{0}]}{[A_{t}]}=kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D%3Dkt)
Where
A0 = initial concentration
At = concentration after time "t"
t = time
k = rate constant
For first order reaction the relation between rate constant and half life is:

Let us calculate k
k = 0.693 / 72 = 0.009625 years⁻¹
Given
At = 0.25 A0

time = 144 years
So after 144 years the sample contains 25% parent isotope and 75% daughter isotopes**
Simply two half lives
Answer:
Ok so, b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...
i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced
ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.
iii. Calculate the standard potential (voltage) of the cell
Look up the reduction potential,
E
⁰
red
, for the reduction half-reaction in a table of reduction potentials
Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,
E
⁰
ox
=
-
E
⁰
red
.
iv. What kind of electrochemical cell is this? Explain your answer.
All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells
1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery
Explanation:
<em><u>Answer and Explanation:</u></em>




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Molar mass of oxygen gas:
O₂ = 16 * 2 = 32.0 g/mol
1 mole O₂ -------------- 32.0
9.05 mole O₂ ---------- ?
Mass = 9.05 * 32.0
Mass = 289.6 g of O₂
hope this helps!
Suppose 110.0 mL of hydrogen gas at STP combines with a stoichiometric amount of fluorine gas and the resulting hydrogen fluoride dissolves in water to form 150.0 mL of an aqueous solution. 0.032 M is the concentration of the resulting hydrofluoric acid.
<h3>What is Balanced Chemical Equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now write the balanced chemical equation
H₂ + F₂ → 2HF
<h3>What is Ideal Gas ?</h3>
An ideal gas is a gas that obey gas laws at all temperature and pressure conditions. It have velocity and mass but do not have volume. Ideal gas is also called perfect gas. Ideal gas is a hypothetical gas.
It is expressed as:
PV = nRT
where,
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant
T = temperature
Here,
P = 1 atm [At STP]
V = 110 ml = 0.11 L
T = 273 K [At STP]
R = 0.0821 [Ideal gas constant]
Now put the values in above expression
PV = nRT
1 atm × 0.11 L = n × 0.0821 L.atm/ K. mol × 273 K

n = 0.0049 mol
<h3>How to find the concentration of resulting solution ? </h3>
To calculate the concentration of resulting solution use the expression

= 0.032 M
Thus from the above conclusion we can say that Suppose 110.0 mL of hydrogen gas at STP combines with a stoichiometric amount of fluorine gas and the resulting hydrogen fluoride dissolves in water to form 150.0 mL of an aqueous solution. 0.032 M is the concentration of the resulting hydrofluoric acid.
Learn more about the Ideal Gas here: brainly.com/question/25290815
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