Plz help me I will mark brainlieat Ques. Write down name and molecular formula of a
1 answer:
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.
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Answer:
:Final block velocity (toward the left)
:Final mass velocity (to the right)
Explanation:
To solve this problem we apply the theory of shocks:
In an elastic shock the kinetic energy and the amount of linear movement or momentum are conserved.
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.
Principle of conservation of the momentum:
m1vi1+m2vi2=m1vf1+m2vf2 Equation 1
Formula to calculate the coefficient of elastic restitution (e):
Equation 2
m1: Block mass
m2: mass of the body that collides with the block
Vi1,vf1: initial, final velocity of the block
Vi2,vf2: initial, final velocity of the body that collides with the block
Of the problem data we know that:
m1=m , m2 = 2m, vi1=v and vi2=0, then, we replace this information in equation (1) :
mv+0=mvf1+2mvf2 we divide by m:
v=vf1+2vf2 Equation (3)
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1,then , we we replace this information in equation (2)
v=vf2-vf1 Equation (4)
vf2=v+vf1 Equation (5)
We replace the equation 5 in the equation (3)
v=vf1+2(v+vf1)
v=vf1+2v+2vf1
-3vf1=v
We replace Vf1=(-1/3)v in the equation (5):
Answer;
:Final block velocity (toward the left)
:Final mass velocity (to the right)