The rate constant for 1st order reaction is
K = (2.303 /t) log (A0 /A)
Where, k is rate constant
t is time in sec
A0 is initial concentration
(6.82 * 10-3) * 240 = log (0.02 /A)
1.63 = log (0.02 /A)
-1.69 – log A = 1.63
Log A = - 0.069
A = 0.82
Hence, 0.82 mol of A remain after 4 minutes.
Answer:
7.23 J
Explanation:
Step 1: Given data
- Mass of graphite (m): 566.0 mg
- Initial temperature: 5.2 °C
- Final temperature: 23.2 °C
- Specific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹
Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)
Q = 7.23 J
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For N₂,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference
0.00 (Non Polar Covalent)
For Na₂O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic)
For CO₂,
E.N of Oxygen = 3.44
E.N of Carbon = 2.55
________
E.N Difference 0.89 (Polar Covalent)
Answer:
The answer to your question is:
a) 0.023 mol of CdS
b) 0.12 mol of MoO3
c) 7.3 mol of AlPO4
Explanation:
a) CdS
MW = 112.4 + 32
= 144 g
144g ------------------ 1 mol
3.28g -------------- x
x = (3.28 x 1) / 144
x = 0.023 mol
b) MoO3
MW = 96 + 3(16) = 144 g
144g ------------------- 1 mol
17.5g ----------------- x
x = (17.5 x 1) / 144
x = 0.12 mol
c) AlPO4
MW = 27 + 31 + 4(16)
MW = 122 g
122 g ------------------ 1mol
890 g ------------------ x
x = (890 x 1) / 122
x = 7.3 mol of AlPO4
Answer:
The artifact is 570 years old. That is, 5.7 × 10² years.
Explanation:
Radioactive decay follows first order reaction kinetics.
Let the initial activity for fresh Carbon-14 be A₀
And the activity at any other time be A
The rate of radioactive decay is given by
dA/dt = - KA
dA/A = - kdt
Integrating the left hand side from A₀ to A₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)
In [(A₀/2)/A₀] = - k t(1/2)
In (1/2) = - k t(1/2)
- In 2 = - k t(1/2)
k = (In 2)/t₍₁,₂₎
t(1/2) is given in the question to be 5.73 × 10³ years
k = (In 2)/5730 = 0.000121 /year
dA/A = - kdt
Integrating the left hand side from A₀ to A and the right hand side from 0 to t
In (A/A₀) = - kt
A/A₀ = e⁻ᵏᵗ
A = A₀ e⁻ᵏᵗ
A = 2.8 × 10³ Bq.
A₀ = 3.0 × 10³ Bq.
2.8 × 10³ = 3.0 × 10³ e⁻ᵏᵗ
0.9333 = e⁻ᵏᵗ
e⁻ᵏᵗ = 0.9333
-kt = In 0.9333
- kt = - 0.06899
t = 0.06899/0.000121 = 570.2 years = 5.7 × 10² years
