Answer:
Explanation:
Given that:
The flow rate Q = 0.3 m³/s
Volume (V) = 200 m³
Initial concentration
= 2.00 ms/l
reaction rate K = 5.09 hr⁻¹
Recall that:







where;







Thus; the concentration of species in the reactant = 102.98 mg/l
b). If the plug flow reactor has the same efficiency as CSTR, Then:
![t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]](https://tex.z-dn.net/?f=t%20_%7BPFR%7D%20%3D%20%5Cdfrac%7B1%7D%7Bk%7D%20%5CBig%20%5B%20In%20%28%20%5Cdfrac%7BC_o%7D%7BC_e%7D%29%20%5CBig%20%5D)
![\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7BPFR%7D%7D%7BQ_%7BPFR%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7Bk%7D%20%5CBig%20%5B%20In%20%28%20%5Cdfrac%7BC_o%7D%7BC_e%7D%29%20%5CBig%20%5D)
![\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7BPFR%7D%7D%7BQ_%7BPFR%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B5.09%7D%20%5CBig%20%5B%20In%20%28%20%5Cdfrac%7B200%7D%7B102.96%7D%29%20%5CBig%20%5D)
![\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7BPFR%7D%7D%7BQ_%7BPFR%7D%7D%20%3D0.196%20%5CBig%20%5B%20In%20%28%201.942%29%20%5CBig%20%5D)





The volume of the PFR is ≅ 140 m³
Answer:
See explanation
Explanation:
The shorthand nuclear reaction equations have been given; the first particle in the parentheses is a reactant particle while the second particle is a product particle. These can now be rewritten as the longhand equations as follows;
238/92U + 4/2 He -------> 241/94Pu + 1/0 n
238/92U + 4/2 He ------> 241/94Pu + 1/0 n
14/7N + 4/2 He------> 17/8O + 1/1 p
56/26Fe + 2 4/2 He----> 60/29Cu + 4/2 He
An isoelectronic series is where all of the ions listed have the same number of electrons in their atoms. When an atom has net charge of zero or neutral, it has equal number of protons and electrons. Hence, it means that the atomic number = no. of protons = no. of electrons. If these atoms become ions, they gain a net charge of + or -. Positive ions are cations. This means that they readily GIVE UP electrons, whereas negative ions (anions) readily ACCEPT electrons. So, to know which of these are isoelectronic, let's establish first the number of electron in a neutral atom from the periodic table:
Na=11; K=19; Rb=37; Cs = 55; Ca=20; S=16; Mg=12; Li=3; Be=4; B=5; C=6, Ar = 18
A. Na⁺: 11-1 = 10 electrons
K⁺: 19 - 1 = 18 electrons
Rb⁺: 37-1 = 36 electrons
B. K⁺: 19 - 1 = 18 electrons
Ca²⁺: 20 - 2 = 18 electrons
Ar: 18 electrons
S²⁻: 16 +2 = 18 electrons
C. Na⁺: 11-1 = 10 electrons
Mg²⁺: 12 - 2 = 10 electrons
S²⁻: 16 +2 = 18 electrons
D. Li=3 electrons
Be=4 electrons
B=5 electrons
C=6 electrons
The answer is letter B.