Consider the following equilibrium:
4NH₃ (g) + 3O₂ (g) ⇄ 2N₂ (g) + 6H₂O (g) + 1531 kJ
The given statement is True, because
According to Le Ch's Principle:
Systems that have attained the state of chemical equilibrium will tend to maintain their equilibrium state.
External factors such as the addition of products and reactants result in the disruption of the equilibrium state.
we expect the system to shift to the direction that offsets the change in concentration.
This results in the state of chemical equilibrium to be reestablished.
Hence, The statement is true,
- The addition of more ammonia (a reactant) would offset the state of equilibrium.
- To restore chemical equilibrium, the system must consume the excess reactants to form more products.
- A shift to favor the products side occurs.
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Answer:
The correct option is (c)
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:
LiCl = 0.492 m
Explanation:
Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.
Our solute is lithium chloride, LiCl.
Our solvent is distilled water.
We do not have the mass of water, but we know the volume, so we should apply density to determine mass.
Density = mass / volume
Density . volume = mass
1 g/mL . 19.7 mL = 19.7 g
We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg
Let's determine the moles of LiCl
0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles
Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m