Students in science class were asked to do an experiment with a can of soda. They placed an unopened can of soda on a digital sc
ale and recorded the initial mass as 340 g. They then opened the soda can and left it open for a day. The following day, the students took the mass of the soda can again and noticed that it was only 332g. They then poured some of the soda into a glass and observed that it was no longer fizzy.
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The compound is (Sulphuric Acid) H2SO4. On reacting with (Sodium Hydroxide) NaOH, it gives (2 Water Molecules/Colored) 2H2O and (1 Sodium Sulfate Molecule/Salt) Na2SO4
H2SO4 + NaOH —> 2H2O (aq.) + Na2SO4 (salt)
The resulted salt/compound (Na2SO4) when reacting with Methyl Orange (MO) is called ”Removal of methyl orange dye and Na2SO4 salt from synthetic wastewater using reverse osmosis (RO)”
The efficiency of reverse osmosis (RO) membranes used for treatment of colored water effluents can be affected by the presence of both salt and dyes.
Concentration polarization of each of the dye and the salt and the possibility of a dynamic membrane formed by the concentrated dye can affect the performance of the RO membrane.
The objective of the current work was to study the effect of varying the Na2SO4 salt and methyl orange (MO) dye concentrations on the performance of a spiral wound polyamide membrane.
The work also involved the development of a theoretical model based on the solution diffusion (SD) mass transport theory that takes into consideration a pressure dependent dynamic membrane resistance as well as both salt and dye concentration polarizations.
Control tests were performed using distilled water, dye/water and salt/water feeds to determine the parameters for the model.
The experimental results showed that increasing the dye concentration from 500 to 1000 ppm resulted in a decrease in the salt rejection at all of the operating pressures and for both feed salt concentrations of 5000 and 10,000 ppm.
Increasing the salt concentration from 5000 to 10,000 ppm resulted in a slight decrease in the percent dye removal. The model’s results agreed well with these general trends.
H2SO4 + NaOH —> 2H2O (aq.) + Na2SO4 (salt)
The resulted salt/compound (Na2SO4) when reacting with Methyl Orange (MO) is called ”Removal of methyl orange dye and Na2SO4 salt from synthetic wastewater using reverse osmosis (RO)”
The efficiency of reverse osmosis (RO) membranes used for treatment of colored water effluents can be affected by the presence of both salt and dyes.
Concentration polarization of each of the dye and the salt and the possibility of a dynamic membrane formed by the concentrated dye can affect the performance of the RO membrane.
The objective of the current work was to study the effect of varying the Na2SO4 salt and methyl orange (MO) dye concentrations on the performance of a spiral wound polyamide membrane.
The work also involved the development of a theoretical model based on the solution diffusion (SD) mass transport theory that takes into consideration a pressure dependent dynamic membrane resistance as well as both salt and dye concentration polarizations.
Control tests were performed using distilled water, dye/water and salt/water feeds to determine the parameters for the model.
The experimental results showed that increasing the dye concentration from 500 to 1000 ppm resulted in a decrease in the salt rejection at all of the operating pressures and for both feed salt concentrations of 5000 and 10,000 ppm.
Increasing the salt concentration from 5000 to 10,000 ppm resulted in a slight decrease in the percent dye removal. The model’s results agreed well with these general trends.
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t = 750 / 2(1.20)
t = 750 / 2.4
t = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t = 750 / 4(1.50)
t = 750 / 6
t = 125 nm
Therefore, the minimum thickness be now will be 125 nm