Answer:
15912 × 10∧-19 KJ
Explanation:
Given data;
frequency of photon = 2.4 × 10^18 1/s.
Planck's constant = 6.63 × 10∧-34 j.s
Energy = ?
Formula:
E = h × ν
E = 6.63 × 10∧-34 j.s × 2.4 × 10^18 1/s
E= 15.912 × 10∧-16 j
now we will convert the joule into kilo joule,
E = 15.912 × 10∧-16 j /1000 = 15.912 × 10∧-19 KJ
The number of mole of Ca reacted is:
4.86 g Ca/ (40.08 g/mol Ca)= 0.121 mol Ca
Because Ca reacted completely with oxygen and there is 2 mol Ca, there is 1 mol O2 reacted.
Total mass of oxygen that reacted is:
0.121 mol Ca* (1mol O2/ 2 mol Ca)* (32 g O2/ 1 mol O2)= 1.94 g O2 reacted.
Hope this would help~
Answer:
5.8μg
Explanation:
According to the rate or decay law:
N/N₀ = exp(-λt)------------------------------- (1)
Where N = Current quantity, μg
N₀ = Original quantity, μg
λ= Decay constant day⁻¹
t = time in days
Since the half life is 4.5 days, we can calculate the λ from (1) by substituting N/N₀ = 0.5
0.5 = exp (-4.5λ)
ln 0.5 = -4.5λ
-0.6931 = -4.5λ
λ = -0.6931 /-4.5
=0.1540 day⁻¹
Substituting into (1) we have :
N/N₀ = exp(-0.154t)----------------------------- (2)
To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:
N = 5.0 μg
N₀ = Unknown
t = 1 day
Substituting into (2) we have
[5/N₀] = exp (-0.154 x 1)
5/N₀ = 0.8572
N₀ = 5/0.8572
= 5.8329μg
≈ 5.8μg
The Chemist must order 5.8μg of 47-CaCO3
