1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
Answer:
Less
Explanation:
Since [Cu(NH3)4]2+ and [Cu(H2O)6]2+ are Octahedral Complexes the transitions between d-levels explain the majority of the absorbances seen in those chemical compounds. The difference in energy between d-levels is known as ΔOh (ligand-field splitting parameter) and it depends on several factors:
- The nature of the ligand: A spectrochemical series is a list of ligands ordered on ligand strength. With a higher strength the ΔOh will be higher and thus it requires a higher energy light to make the transition.
- The oxidation state of the metal: Higher oxidation states will strength the ΔOh because of the higher electrostatic attraction between the metal and the ligand
A partial spectrochemical series listing of ligands from small Δ to large Δ:
I− < Br− < S2− < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < CH3CN < NH3 < NO2− < PPh3 < CN− < CO
Then NH3 makes the ΔOh higher and it requires a higher energy light to make the transition, which means a shorter wavelength.
Answer:
6.67 moles
Explanation:
Given that:-
Moles of hydrogen gas produced = 10.0 moles
According the reaction shown below:-
3 moles of hydrogen gas are produced when 2 moles of aluminium undergoes reaction.
Also,
1 mole of hydrogen gas are produced when 
moles of aluminium undergoes reaction.
So,
10.0 moles of hydrogen gas are produced when 
moles of aluminium undergoes reaction.
<u>Moles of Al needed = moles = 6.67 moles</u>
Answer:
Elements can be described by various properties, and identified by their boiling and melting points. For example, gold melts at
Elements can be described by various properties, and identified by their boiling and melting points. For example, gold melts at 1,064ºC and boils at 2,856ºC. Does boiling point depend on the mass present?
A. No; chemical properties stay the same regardless of mass.
B. No; physical properties stay the same regardless of mass.
C. Yes; physical properties can change when mass increases or decreases.
D. No; qualitative properties like boiling point stay the same regardless of mass.
Explanation:
<span>The answer to your question is the 3rd option </span>
