Explanation:
If a large photon strikes the surface, that has enough strength to take out an electrode, which will then travel to the positive side since it is negative. Current is flowing at this stage. Since the reduced photons will be unable to distinguish between atoms, no power can pass.
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Answer:</h3>
812 kPa
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Explanation:</h3>
- According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
- Mathematically,
At varying pressure and volume;
P1V1=P2V2
In this case;
Initial volume, V1 = 2.0 L
Initial pressure, P1 = 101.5 kPa
Final volume, V1 = 0.25 L
We are required to determine the new pressure;
Replacing the known variables with the values;
= 812 kPa
Thus, the pressure of air inside the balloon after squeezing is 812 kPa
Answer:
V = 42.6 L
Explanation:
Given data:
Number of moles of Cl₂ = 1.9 mol
Temperature and pressure = standard
Volume occupy = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm × V = 1.9 mol ×0.0821 atm.L /mol.K × 273.15 k
V = 42.6 atm.L / 1 atm
V = 42.6 L
Answer: D
Explanation:
A reducing agent is a species that reduces other compounds, and is thereby oxidized. The whole compound becomes the reducing agent. In other words, of a compound is oxidized, then they are the reducing agent. On the other hand, if the compound is reduced, it is an ozidizing agent.
Since we have established that a reducing agent is the compound being oxidized, we know that A is not our answer. An oxidized compound is losing electrons. Choice A states exactly this.
For B, this is true as we have established this already.
C is also correct. Since a reducing agent loses electrons, it becomes more positive. This makes the oxidation number increase.
D would be our correct answer. It is actually a good oxidizing agent is a metal in a high oxidation state, such as Mn⁷⁺.
60 neutrons are in silver
