N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) =
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).
I’m assuming it’s type one (most likely) diabetes!
The volume of chlorine required is 7.71 L.
The reaction between phosphorus and chlorine is:
2P + 5Cl₂→ 5PCl₅
Therefore, 2 moles of P requires 5 moles of chlorine to react with it.
Given mass of P =3.39 g
Molar mass of P=30.97 g/mol
No. of moles of P = given mass/ molar mass = 3.39 / 30.97 = 0.109 moles
2 moles of P requires 5 moles of chlorine
0.109 moles of P will require 0.109 x 5/2 = 0.2725 moles of chlorine
According to ideal gas equation
PV=nRT
2.04 x V = 0.2725 x 0.0821 x 703
V = 0.2725 x 0.0821 x 703 / 2.04
V = 7.71L
Learn more about ideal gas equation here:
brainly.com/question/3637553
#SPJ4
Explanation:
A point of temperature at which both solid and liquid state of a substance remains in equilibrium without any change in temperature then this temperature is known as melting point.
For example, melting point of water is 
. So, at this temperature solid state of water and liquid state are present in equilibrium with each other.
Therefore, when a 100 g of given pure metal in solid state is heated at its exact melting point which is 
then some of the solid will change into liquid state but the temperature will remains the same.
Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J
