Ultraviolet as ultraviolet light waves are around 420 nm like stated in the photo
Answer:
26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃
Explanation:
To determine the number of moles of O₂ that are needed to react completely with 35.0 mol of FeCl₃, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), and rule of three as follows: if 4 moles of FeCl₃ react with 3 moles of O₂, 35 moles of FeCl₃ with how many moles of O₂ will it react?

moles of O₂= 26.25 ≅ 26.3
<u><em>26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃</em></u>
Answer:
the smallest particle of a chemical element that can exist.
Explanation:
Noble Gas. Metals have 1 or 2 Valence Electrons. Halogens have 7 Valence Electrons. Semi-Metals can have different amounts.
Answer:
The new pressure will be 0.225 kPa.
Explanation:
Applying combined gas law:

where,
are initial pressure and volume at initial temperature
.
are final pressure and volume at initial temperature
.
We are given:



Putting values in above equation, we get:


Hence, the new pressure will be 0.225 kPa.