<span>A. the number of perching birds present</span>
We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.
The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.
Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-
N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)
According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, moles = 0.375 moles of nitrogen is required.
Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.
Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.
Answer:
The answer to your question is given below
Explanation:
1. When two hydrogen atoms pair together, they form a single bond to achieve complete duet.
Hydrogen atom has 1 shell and 1 valence electron and they contribute it to form bond, thereby obtaining a stable duet configuration
H + H —> H–H —> H2
2. When two halogen atom pair together, they form a single bond to achieve a complete octet which is stable. For example, Cl has 7 electrons in its outermost shell and need 1 more electron to complete it octet. Each Cl will contribute 1 electron each to form a single bond..
Cl + Cl —> Cl–Cl —>Cl2
3. When two oxygen atoms pair together, they form double bond to achieve a complete octet which is a stable configuration.
Oxygen has 6 electrons in its outermost shell and needs 2 more electrons to complete its octet configuration. Each oxygen atom will contribute 2 electrons each to form a double in order to complete it octet configuration.
O + O —> O=O —>O2
4. When two nitrogen atom pair together, they form a triple bond to achieve a complete octet which is stable. Nitrogen has 5 electrons in its outermost shell and needs 3 more electrons to complete it octet configuration. Each nitrogen atom will contribute 3 electrons each to form a triple bond in order to complete its octet configuration.
N + N —> N≡N —> N2.