answer: dispersed from the liquid so cold air can take its place
Soft light (answer) is the filament between (2700k-3000k).
The higher the kelvin number the whiter the light.
3500k-4100k is bright white/cool white
5000k-6500k is daylight
\and those are the three primary colors of color temperature
The replacing of sodium hydroxide with potassium hydroxide
(KOH) to the reaction will least affect the organic product that forms.
Potassium hydroxide is an
inorganic compound with the formula KOH, and is commonly called caustic potash. Along with sodium hydroxide, this colorless
solid is a prototypical strong base.
D. a compound can only be separated into its components by chemical means
Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
![\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20KClO%7D_%7B3%7D%20%3D%20%5Ctext%7B100.0%20mol%20KCl%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mol%20KClO%7D_%7B3%7D%7D%7B%5Ctext%7B2%20mol%20KCl%7D%7D%20%3D%20%5Ctext%7B100.0%20mol%20KClO%7D_%7B3%7D)
(ii) From O₂
![\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20KClO%7D_%7B3%7D%20%3D%20%5Ctext%7B100.0%20mol%20O%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mol%20KClO%7D_%7B3%7D%7D%7B%5Ctext%7B3%20mol%20O%7D_%7B2%7D%7D%20%3D%20%5Ctext%7B66.67%20mol%20KClO%7D_%7B3%7D)
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
![\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20used%7D%20%3D%20%5Ctext%7B100.0%20mol%20O%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mol%20KCl%7D%7D%7B%5Ctext%7B3%20mol%20O%7D_%7B2%7D%7D%20%3D%20%5Ctext%7B66.67%20mol%20KCl%7D)
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol