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.
Answer:
E = 2.7 x 10¹⁶ J
Explanation:
The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:

where,
E = Energy Released = ?
m = mass of material reduced = 0.3 kg
c = speed of light = 3 x 10⁸ m/s
Therefore,

<u>E = 2.7 x 10¹⁶ J</u>
Answer:
Part a)
A = 0.066 m
Part b)
maximum speed = 0.58 m/s
Explanation:
As we know that angular frequency of spring block system is given as

here we know
m = 3.5 kg
k = 270 N/m
now we have


Part a)
Speed of SHM at distance x = 0.020 m from its equilibrium position is given as



Part b)
Maximum speed of SHM at its mean position is given as


Answer:

Explanation:
As we know that moment of force is given as

now we have


now from above formula we have

here we know that

so we have


Answer:
Stretch can be obtained using the Elastic potential energy formula.
The expression to find the stretch (x) is 
Explanation:
Given:
Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.
To find: Elongation in the spring (x).
We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).
The formula to find EPE is given as:

Rewriting the above expression in terms of 'x', we get:

Example:
If EPE = 100 J and spring constant, k = 2 N/m.
Elongation or stretch is given as:

Therefore, the stretch in the spring is 10 m.
So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.