Answer:
7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.
Explanation:
First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.
According to the given question we have to prepare 0.100 M solution
1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4
1 ml of solution contain 14.208÷1000= 0.014 gram
0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.
So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.
Answer:
4 moles, 160 g
Explanation:
The formula for the calculation of moles is shown below:
For 
:-
Mass of = 196 g
Molar mass of = 98 g/mol
The formula for the calculation of moles is shown below:
Thus,
According to the given reaction:
1 mole of sulfuric acid reacts with 2 moles of NaOH
So,
2 moles of sulfuric acid reacts with 2*2 moles of NaOH
Moles of NaOH must react = 4 moles
Molar mass of NaOH = 40 g/mol
<u>Mass = Moles*molar mass = 
= 160 g</u>
Answer:
Conductivity meter
Explanation:
A conductivity meter is normally used to measure the amount of electrical current or conductance in a solution. Conductivity is most useful in determining the overall health of a natural water body.
A pH paper is used to determine the pH of a solution. This is done by dipping part of the paper into a solution of interest and watching the color change. The pH paper comes in a color-coded scale indicating the pH that something has when the paper turns a certain color.
An indicator is an organic compound that changes its colour depending on the pH of the solution.
Since neutralization reaction can only be monitored by monitoring the pH of the solution, a conductivity meter cannot be used to monitor the progress of a neutralization reaction since it does not monitor the change in pH of the system under study.
<u>Answer:</u> C) be hypertonic to Tank B.
<u>Explanation: </u>
<u>
The ability of an extracellular solution to move water in or out of a cell by osmosis</u> is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the <u>total concentration of all the solutes in the solution.
</u>
Three terms (hypothonic, isotonic and hypertonic) are used <u>to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane</u>. When we use these terms, we only take into account solutes that can not cross the membrane, which in this case are minerals.
- If the liquid in tank A has a lower osmolarity (<u>lower concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypotonic with respect to the latter.
- If the liquid in tank A has a greater osmolarity (<u>higher concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypertonic with respect to the latter.
- If the liquid in tank A has the same osmolarity (<u>equal concentration of solute</u>) as the liquid in tank B, the liquid in tank A would be isotonic with respect to the latter.
In the case of the problem, option A is impossible because the minerals can not cross the membrane, since it is permeable to water only. There is no way that the concentration of minerals decreases in tank A, so <u>the solution in this tank can not be hypotonic with respect to the one in Tank B. </u>
Equally, both solutions can not be isotonic and neither we can say that the solution in tank A has more minerals that the one in tank B because the liquid present in tank B is purified water that should not have minerals. Therefore, <u>options B and D are also not correct.</u>
Finally, the correct option is C, since in the purification procedure the water is extracted from the solution in tank A to obtain a greater quantity of purified water in tank B. In this way, the solution in Tank A would be hypertonic to Tank B.
