Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value
<span>
The half-life for the chemical reaction is 29,2 s and is
independent of initial concentration.
c</span>₀
- initial concentration the reactant.
c - concentration of the reactant remaining
at time.
t = 29,2 s.<span>
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s</span> = 0,0237 1/s.<span>
ln(c/c</span>₀) = -k·t₁.<span>
ln(1/16 </span>÷ 1) = -0,0237 1/s ·
t₁.
t₁ = 116,8 s.
ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹
According to the balanced equation,
ΣδΗ(bond breaking) = N ≡ N x 1 + Cl - Cl x 3
= 945 + 3(242)
= 1671 kJ mol⁻¹
ΣδΗ(bond making) = N - Cl x 3 x 2
= 192 x 6
= 1152 kJ mol⁻¹
δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
= 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
= 519 kJ mol⁻¹
Answer:
Its in the Explanation
Explanation:
Here's what I got.
Aluminium-27 is an isotope of aluminium characterized by the fact that is has a mass number equal to
27
.
Now, an atom's mass number tells you the total number of protons and of neutrons that atom has in its nucleus. Since you're dealing with an isotope of aluminum, it follows that this atom must have the exact same number of protons in its nucleus.
The number of protons an atom has in its nucleus is given by the atomic number. A quick looks in the periodic table will show that aluminum has an atomic number equal to
13
.
This means that any atom that is an isotope of aluminum will have
13
protons in its nucleus.
Since you're dealing with a neutral atom, the number of electrons that surround the nucleus must be equal to the number of protons found in the nucleus.
Therefore, the aluminium-27 isotope will have
13
electrons surrounding its nucleus.
Finally, use the known mass number to determine how many neutrons you have
mass number
=
no. of protons
+
no. of neutrons
no. of neutrons
=
27
−
13
=
14
Your welcome :)
