Answer:
C:to the right side of the periodic table, and it is given
the suffix -ide.
Answer:
chemical potential energy - It mainly has chemical potential energy, this is really a type of electrical potential energy stored in the chemical bonds of the molecules
Explanation:
You can give the other person brainliest, I don't need it! :)
Answer:
The limiting reactant is acetic acid. All 125 g will react.
Explanation:
1. Assemble the information
We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
Mᵣ: 60.05 78.00
3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O
Mass/g: 125 275
2. Calculate the moles of each reactant

3. Calculate the moles of (CH₃COO)₃Al from each reactant


Answer:
A
Explanation:
Ionic compounds are solids when at room temperature