Cytotoxic is a characteristic of a medicine that bears the ability to cure cancer. The timing of the medication is very important because if not used properly, the substance or the drug may damage also even the healthy cells. Thus, it is just right to caution everyone, especially those who are involved, with the right usage of the drug.
Answer: The concentration of
ions in the resulting solution is 1.16 M.
Explanation:
To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

where,
are the n-factor, molarity and volume of the 
are the n-factor, molarity and volume of the 
We are given:
Putting all the values in above equation, we get

The concentration of
ions in the resulting solution will be same as the molarity of solution which is 1.16 M.
Hence, the concentration of
ions in the resulting solution is 1.16 M.
<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass.
For this, we look at the atomic masses of the elements present in the compound.
Cu has an atomic mass of 63.546 amu
Fe has 55.845 amu
and S has 36.065 amu
Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu
So we have not established the mass of the compound in amus
63.546 + 55.845 + 72.13 = 191.521
That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845
So to get the percentage, or fraction of iron, we take 55.845 / 191.521
Which comes out to 29.15% by mass
Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>
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