To solve this, we can use two equations.
t1/2 = ln 2 / λ = 0.693 / λ
where, t1/2 is half-life and λ is the decay constant.
t1/2 = 10 min = 0.693 / λ
Hence, λ = 0.693 / 10 min - (1)
Nt = Nο e∧(-λt)
Nt = amount of atoms at t =t time
Nο= initial amount of atoms
t = time taken
by rearranging the equation,
Nt/Nο = e∧(-λt) - (2)
From (1) and (2),
Nt/Nο = e∧(-(0.693 / 10 min) x 20 min)
Nt/Nο = 0.2500
Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%
= (Nt/Nο ) x 100%
= 0.2500 x 100%
= 25.00%
Hence, Percentage of remaining nuclei is 25.00%
It’s B because of the quest
Its phosphorus (P)In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.
For the first one the pattern is multiply the previous number by five as you see 1 x 5 = 5 and so on. To keep adding to it you would do
125 x 5 = 625 625 x 5 = 3125 3125 x 5 = 15625
Now for the second one the pattern is divide the previous number by three as you can see 2187 / 3 = 729 and so on. To keep going you would
81 / 3 = 27 27 / 3 = 9 9 / 3 = 3
I hope this helps you and if you have anymore questions i'll be glad to answer them.
Answer is: mass of water is 56.28 grams.
Chemical reaction: 2H₂O → 2H₂ + O₂.
m(O₂) = 50.00 g.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 50 g ÷ 32 g/mol.
n(O₂) = 1.5625 mol.
From chemical reaction: n(O₂) : n(H₂O) = 1 : 2.
n(H₂O) = 2 · 1.5625 mol.
n(H₂O) = 3.125 mol.
m(H₂O) = n(H₂O) · M(H₂O).
m(H₂O) = 3.125 mol · 18.01 g/mol.
m(H₂O) = 56.28 g.
