The definition of dilute is "make (a liquid) thinner or weaker by adding water or another solvent to it." Now, this may make you think that the beaker with three scoops is the most dilute, but it's not. In this case, it depends on the salt to water ratio. Let's say each beaker contains five parts water. The first beaker has a ratio of 1/5. The second had a ratio of 2/5. The third has a ratio of 3/5. To find which has the most water compared to the others, I'll use equal to make the numerator (The amount of salt) seemingly equal each time. Just a warning, this strategy doesn't work every time. Now, if we make the numerators the same, that means which ever denominator is the highest will be the most dilute solution. Let's make each numerator equal to six, as each number (1, 2, and 3) go into six.
1/5 = 6/30
2/5 = 6/15
3/5 = 6/12
I got these numbers by dividing six (What we want the numerator to be) by each current numerator, and then multiplying the quotient (The answer of a division problem) by both sides of the fraction. Since the first beaker has the highest denominator, we know that it is the most dilute.
mark brainliest ;)
I thought you were going to ask for the resistance of the unknown
series resistor. Since you only want the equivalent resistance of the
circuit, you don't even need to know the resistance of the lamp.
I = E / R
Current through the circuit = (voltage of the battery) / (circuit resistance).
0.5 = (12) / R
Multiply each side by 'R' : (0.5) R = 12
Multiply each side by 2 : <em>R = 24 ohms</em>
(Since the resistance of the lamp is 10 ohms, the
unknown series resistor is the other 14 ohms.)
T&hey are heavier I believe and need a bigger engine to make it
hope i helped
Have a great day
Visible light is the only light visible to the human eye.
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.