<span><u><em>Answer:</em></u>
All of the above
<u><em>Explanation:</em></u>
Vertebrates are a class of creatures falling under kingdom "<u>Animalia</u>" that are characterized by the presence of an internal skeleton composed of bones.
<u>Vertebrates are characterized by the following:</u>
1- presence of internal skeleton
2- developed brain
3- the presence of an advanced nervous system connected to the brain
4- presence of muscles that allow movement
5- protective skin
6- circulation of blood in the bodies in the vessels
Comparing the mentioned characteristics with the options given, we will find that the most suitable answer is: <u>"all of the above"</u>.
Hope this helps :)</span>
The empirical formula is K₂O.
The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.
The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.
So, our job is to calculate the <em>molar ratio</em> of K to O.
Step 1. Calculate the <em>moles of each element
</em>
Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K
Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0
Step 2. Calculate the <em>molar ratio of each elemen</em>t
Divide each number by the smallest number of moles and round off to an integer
K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1
Step 3: Write the <em>empirical formula
</em>
EF = K₂O
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