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navik [9.2K]
3 years ago
10

Differences between alkalis, bases and acids? (answer for many points :) )

Chemistry
1 answer:
Rudik [331]3 years ago
3 0

Explanation:

protons.

An acid can also be thought of as a chemical that can neutralize a base. Similarly, a base can neutralize an acid.

Acids turn litmus paper red, while bases make litmus paper turn blue.

Some examples of acids are sulfuric acid, hydrochloric acid, nitric acid, and so on. Some examples of bases are sodium hydroxide, potassium hydroxide, and so on.

Acids generally taste sour, while bases have a bitter taste.

Alkalis are the bases that are water-soluble, which means that they dissolve in water. In other words, not all bases are water-soluble, and only the water-soluble bases are known as alkalis. An example of an alkali is sodium hydroxide. It is a base because it can neutralize an acid, and because it is water-soluble, it is an alkali. An example of a base that is not alkali is copper oxide. This chemical can neutralize an acid, but it is insoluble in water.

In other words, all alkali are bases but not all bases are alkalis.

Also, an alkali has a hydroxide group, while a base has an oxide group in it.

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Cerrena [4.2K]

Answer:

x_H=0.882

x_N=0.118

Explanation:

In this reactor, oleic and linoleic acid react with hydrogen to form stearic acid. This reactions can be represented by:

Oleic: C_{18}H_{34}O_2 (l) + H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Linoleic: C_{18}H_{32}O_2 (l) + 2 H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Having this reactions in mind, the first thing is to determine the moles of hydrogen required:

<u>Base of caculation: 1 mol of sunflower oil</u>

For oleic acid: n_{Holeic}=\frac{1 mol H_2}{1 mol oleic}*\frac{0.27 mol oleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{90.45 mol H_2}{hr}

For linoleic acid: n_{Hlinoleic}=\frac{2 mol H_2}{1 mol linoleic}*\frac{0.59 mol linoleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{395.3 mol H_2}{hr}

n_{Htotal}=\frac{90.45 mol H_2}{hr}+\frac{395.3 mol H_2}{hr}

n_{Htotal}=frac{485.75 mol H_2}{hr}

Applying the excess:

n_{Htotal}=frac{485.75 mol H_2}{hr}*1.65=801.48 mol

Nitrogen: n_N= 801.48 mol*\frac{0.05 mol N}{0.95 mol}

n_N= 42.2 mol N

<u>After the reactions</u>:

n_H=801.48 mol-485.75mol=315.73 mol

and the nitrogen is inert.

Purge stream:

n_total=42.2+315.73 mol=357.93 mol

x_H=\frac{315.73mol}{357.93mol}=0.882

x_N=\frac{42.2mol}{357.93mol}=0.118

4 0
3 years ago
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