(i) We start by calculating the mass of sugar in the solution:
mass of sugar = concentration × solution mass
mass of sugar = 2.5/100 × 500 = 12.5 g
Then now we can calculate the amount of water:
solution mass = mass of sugar + mass of water
mass of water = solution mass - mass of sugar
mass of water = 500 - 12.5 = 487.5 g
(ii) We use the following reasoning:
If 500 g solution contains 12.5 g sugar
Then X g solution contains 75 g sugar
X=(500×75)/12.5 = 3000 g solution
Now to get the amount of solution in liters we use density (we assume that is equal to 1):
Density = mass / volume
Volume = mass / density
Volume = 3000 / 1 = 3000 liters of sugar solution
The NaOH will be used What titrant to titrate the 0. 02 m hcl phenol red solution.
Acid-base titrations may be the most typical titrations, although there are numerous more forms as well. Take a look at this illustration where sodium hydroxide is used to titrate a sample of hydrochloric acid (HCl) (NaOH). The titrant (NaOH), which is added gradually throughout the duration of the titration, has been added to the unknown solution.
Titrants are solutions with known concentrations that are added to solutions whose concentrations must be determined. The solution for whom the concentration needs to be determined is known as a titrant as well as analyte.
Therefore, the NaOH will be used as a titrant to titrate the 0. 02 m hcl phenol red solution.
To know more about titrant
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I believe it is spontaneous generation
Answer: Option (E) is the correct answer.
Explanation:
Since, the conductor is hollow which means that it is opened on both the ends. Hence, when a small uncharged metal ball is passed through it with the help of a silk thread then due to the presence of this insulating thread the ball will not come directly in contact with the charged rod.
As a result, there will occur no formation of opposite charge on the metal ball. Therefore, the ball will remain uncharged in nature.
Thus, we can conclude that after the given ball is removed, it will have no appreciable charge.
