The molar solubility of calcium sulfide in a 0.222 M sodium sulfide solution is:________-
1 answer:
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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.
There are one antibonding molecular orbitals present in molecular orbital model of c.
The cyclobutadiene has a pi system comprised of four individual atomic p - orbital and thus should have a four pi molecular orbitals. The compound is the prototypical antiaromatic hydrocarbon with 4 - electrons . Its rectangular structure is the result of jahn teller reaction which disorder the molecule and lowers its symmetry , converting the triplet to a singlet ground state. It is a small annulene . The delocalisation energy of the
electrons of the cyclobutene is predicted to be zero .
To learn more about antibonding molecular orbitals click here
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Answer:
3m/s²
Explanation:
Given parameters:
Mass of object = 3.2kg
Force to the right = 16.3N
Force to the left = 6.7N
Unknown:
Acceleration of the object = ?
Solution:
To solve this problem, we use newtons second law of motion;
Net force = mass x acceleration
Net force on object = Force to the right - Force to the left
Net force = 16.3N - 6.7N = 9.6N
So;
9.6 = 3.2 x a
a = = 3m/s²