Answer:
Gaseous at standard conditions.
Explanation:
Hello,
In this case, for the given normal boiling point of oxygen, we can notice those are conditions far away from the standard conditions which are 25 °C. In such a way, since the standard temperature is by far higher than the normal boiling point, the most likely state of a sample of pure oxygen would be gaseous since the molecules will move based on the kinetic theory of gases due to the relatively high temperature which make them be separated and moving all around.
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Answer:
A. manganese change from +7 to +2
Explanation:
Reduction means a gain of electrons, which means the oxidation number will go down or the charge goes down basically or become more negative charge. On the left, Mn has +7 in MnO4-. On the right, it has 2+, so it has been reduced.
Answer:
Average atomic mass = 63.62 amu
Explanation:
Average atomic mass is the weighted average of the individual isotopes.
And thus the weighted average =
(0.6917×63+0.3083×65)⋅amu = ?? amu.
Note that we have not accounted for the less massive copper isotopes that would decrease this mass to the accepted value of
63.55 amu.
Hope this helps! Good luck!
The final volume of the gas is 73.359 mL
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Given :
A sample gas has an initial volume of 72.0 mL
The work done = 141.2 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since, the process does 141.2 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm; we have
External Pressure P
:
= 783 torr × 
= 1.03 atm
The work done W = 
The change in volume ΔV= 
ΔV = 
ΔV = 
ΔV = 0.001359 L
ΔV = 1.359 mL
The initial volume = 72.0 mL
The change in volume V is ΔV = V₂ - V₁
- V₂ = - ΔV - V₁
multiply both sides by (-), we have:
V₂ = ΔV + V₁
= 1.359 mL + 72.0 mL
= 73.359 mL
Therefore, the final volume of the gas is 73.359 mL .
Learn more about volume here:
brainly.com/question/27100414
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For the first one I think it’s A, and I think the second one is C. Those seem most logical.
