Answer:
Here's what I find.
Explanation:
An indicator is usually is a weak acid in which the acid and base forms have different colours. Most indicators change colour over a narrow pH range.
(a) Litmus
Litmus is red in acid (< pH 5) and blue in base (> pH 8).
This is a rather wide pH range, so litmus is not much good in titrations.
However, the range is which it changes colour includes pH 7 (neutral), so it is good for distinguishing between acids and bases.
(b) Phenolphthalein
Phenolphthalein is colourless in acid (< pH 8.3) and red in base (> pH 10).
This is a narrow pH range, so phenolphthalein is good for titrating acids with strong bases..
However, it can't distinguish between acids and weakly basic solutions.
It would be colourless in a strongly acid solution with pH =1 and in a basic solution with pH = 8.
(c) Other indicators
Other acid-base indicators have the general limitations as phenolphthalein. Most of them have a small pH range, so they are useful in acid-base titrations.
The only one that could serve as a general acid-base indicator is bromothymol blue, which has a pH range of 6.0 to 7.6.
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
Electronnegativity increase because the number of charges on the nucleus increases. Which attracts the bonding amount of electrons more.
Answer:
The main purpose of the expedition was to conduct a hydrographic survey of the coasts of the southern part of South America.
