Answer:
3. crystal habit and cleavage.
Explanation:
Crystal habit is a distinctive characteristic which is shown in its general shape, crystallographic forms, how developed each form is.
Cleavage is the ability of a mineral to break in smooth planes parallel to zones of weak bonding. Cleavage in three directions at right angles (90o). Cubic cleavage. Cleavage in three directions not at right angles (120o and 60o).
<u>Answer:</u>
<u>For A:</u> The
for the given reaction is 
<u>For B:</u> The
for the given reaction is 1642.
<u>Explanation:</u>
The given chemical reaction follows:

The expression of
for the above reaction follows:

We are given:

Putting values in above equation, we get:

Hence, the
for the given reaction is 
Relation of
with
is given by the formula:

where,
= equilibrium constant in terms of partial pressure = 
= equilibrium constant in terms of concentration = ?
R = Gas constant = 
T = temperature = 500 K
= change in number of moles of gas particles = 
Putting values in above equation, we get:

Hence, the
for the given reaction is 1642.
An acid-base indicator is used to identify the presence of an acid or base. These indicators exhibit different colors depending on the solution that they are in. They are especially useful when completing titrations to determine the molarity of an unknown substance and is denoted as option D.
<h3>What is Titration?</h3>
This is done in the laboratory and involves the slow addition of one solution of a known concentration to a known volume of another solution of unknown concentration.
Acid-base indicators are used to determine the presence of an acid or base in a solution which is based on the colors seen when performing the chemical reaction.
It is used to calculate the the molarity of an unknown substance through the knowledge of the other parameters which is therefore the reason why option D was chosen as the most appropriate choice.
Read more Acid-base indicators here brainly.com/question/2815636
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Answer:
A. The increase in the mass of the magnesium oxide was due to oxygen atoms in the air.
Explanation:
Burning occurs in the presence of oxygen. A chemical combination occurs between Mg and O in that the atom of magnesium attracts one another.
Mg + O₂ → MgO
Starting with the magnesium, on reacting with oxygen a new compound forms. This is why there is mass increase in the MgO compared to the starting material.