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The Lewis structure of
Diimide (N₂H₂) is shown below.
In this molecule two Nitrogen atoms attached to each other through a
double bond are further attached to one one Hydrogen atom. Also, each Nitrogen atom carries one
non-binding electron pair (
Lone Pair) (Highlighted RED).
Result: Option-C (<span>each nitrogen has one nonbinding electron pair) is the correct answer.</span>
Answer:
H₃PO₄/H₂PO₄⁻ and HCO₃⁻/CO₃²⁻
Explanation:
An acid is a proton donor; a base is a proton acceptor.
Thus, H₃PO₄ is the acid, because it donates a proton to the carbonate ion.
CO₃²⁻ is the base, because it accepts a proton from the phosphoric acid.
The conjugate base is what's left after the acid has given up its proton.
The conjugate acid is what's formed when the base has accepted a proton.
H₃PO₄/H₂PO₄⁻ make one conjugate acid/base pair, and HCO₃⁻/CO₃²⁻ are the other conjugate acid/base pair.
H₃PO₄ + CO₃²⁻ ⇌ H₂PO₄⁻ + HCO₃⁻
acid base conj. conj.
base acid
Answer:
39.1-32.5 and you will find your answer it always like that, you subtract your starting point from your ending point
Explanation:
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).