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
Answer: Benzene is less reactive than methylbenzoate and more reactive than Nitrobenzene
Explanation:
This is because the methyl group on the benzene ring is an electron donating group leading to the activation of the ring and subsequently leading to more canonical resonance structure at the intermediate stage of the reaction enhancing the faster reactivity
However for the Nitrobenzene the nitro group is an electron withdrawing group leading to a slower activation and less resonance canonical structure at the reaction intermediate leading to a slower reaction than the reaction of benzene without the nitro group
Answer:
a compound,typically a crystalline one,in which water molecules are chemically bound to another compound or an element
Given:
P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?
Assume that the gas ideally, thus, we can use the ideal gas equation:
PV = nRT
where R = 0.0821 L atm/mol K
123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T
solve for T
T = 295.72 K<span />
