NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete
<h3>What is a limiting reagent?</h3>The reactant that is entirely used up in a reaction is called as limiting reagent.
The reaction:
→
Moles of nitrogen monoxide
Molecular weight: =30g/mol
Moles of hydrogen
Molecular weight: =30g/mol
Hydrogen gas is in excess.
NO is the limiting reagent.
The amount of the excess reagent remains after the reaction is complete.
(2.9 mol- 0.73 mol NO x
) x
4.34 g
Learn more about limiting reagents here:
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Answer: Heat of vaporization is 41094 Joules
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
where,
= initial pressure at 429 K = 760 torr
= final pressure at 415 K = 515 torr
= enthalpy of vaporisation = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 429 K
= final temperature = 515 K
Now put all the given values in this formula, we get
Thus the heat of vaporization is 41094 Joules
Answer:
[Ar] 4s² 3d⁵ or 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
Explanation:
Manganese electronic configuration is [Ar] 3d⁵ 4s². Manganese has 5 electrons in its 3d shell and all five electrons and unpaired maintaining parallel spin since they must obey hund's rule.
Generally electrons first enter 4s² shell filling it first before entering the 3d shell due to the lower energy 4s orbital has compared to 3d orbital. Filling of the 3d orbital must be filled parallel first before pairing begins. However in the case of manganese, it only has five electrons in its 3d orbital which will only fill it with single spins.
Due to the fact that manganese has five electrons in its 3d orbital its highly magnetic and its classified as a "ferromagnetic" substance.
Note when manganese forms a bond, it loses electrons from its 4s orbital too first to form Mn(ii) or combined from both 4s and 3d to form Mn(iii), (iv) (vi) etc.
Attached is a diagram showing how manganese atoms are arranged in its shells
40 ppm of NaOH
<u>Explanation:</u>
ppm is nothing but the parts per million and it is represented as 1 mg solute per kg of solution.
Here, number of moles of solute (NaOH) can be found as 0.001 mol / L, so we can find the amount of solute as,
Moles =
0.001 mol / L =
0.001 × 40 = 0.04 g of solute
1 g = 1000 mg
0.04 g = 40 mg
In 1 kg of solution 40 mg of solvent is present.
So the concentration of NaOH is 40 ppm.