The answer is letter a. Hill terrain, hill slope, and hill elevation
Answer:
No.of moles of C is , n = mass/molar mass = 75.46 g / 12 (g/mol) = 6.3 moles No.of moles of H is , n' = mass/molar mass = 4.43 g / 1.0(g/mol) = 4.43 moles No.of moles of O is , n'' = mass/molar mass = 20.10 g / 16(g/mol) =1.25 moles Ratio to the no.of moles of C,H& O is 6.3 : 4.43 : 1.25 In the simple integer ratio is ( 6.3/1.25) : ( 4.43/1.25) : (1.25/1.25) 5.04 :3.5 : 1
Explanation:
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
Answer:
There are two types of hydrocarbons: aliphatic and aromatic. The three types of aliphatic hydrocarbons are alkanes, alkenes, and alkynes. Aromatic hydrocarbons include benzene. Overall, examples of hydrocarbons are methane, ethane, propane, and butane.
Like if you imagine making something you create a hypothesis about was to make that happen
