Answer:
<h2>1.89 atm</h2>
Explanation:
The new volume can be found by using the formula for Boyle's law which is

Since we are finding the new volume

From the question we have

We have the final answer as
<h3>1.89 atm</h3>
Hope this helps you
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:
A_________________________
Explanation:
Chromosomal DNA is packaged inside microscopic nuclei with the help of histones. These are positively-charged proteins that strongly adhere to negatively-charged DNA and form complexes called nucleosomes. Each nuclesome is composed of DNA wound 1.65 times around eight histone proteins.
Answer:
1.32×10²⁵ atoms of sulfate are contained in 22 units of it
Explanation:
1 unit = 1 mol
Al₂(SO₄)₃ → Aluminum sulfate
As 1 unit = 1 mol, 1 unit has 6.02×10²³ atoms of aluminum sulfate.
Let's make a rule of three:
1 unit of Al₂(SO₄)₃ contains 02×10²³ atoms
Then, 22 units of Al₂(SO₄)₃ must contain (22 . 6.02×10²³) / 1 = 1.32×10²⁵ atoms