Hey there!
C₉H₂O + O₂ → CO₂ + H₂O
First let's balance the C.
There's 9 on the left and 1 on the right. So, let's add a coefficient of 9 in front of CO₂.
C₉H₂O + O₂ → 9CO₂ + H₂O
Next let's balance the H.
There's 2 on the left and 2 on the right. This means it's already balanced.
C₉H₂O + O₂ → 9CO₂ + H₂O
Lastly, let's balance the O.
There's 3 on the left and 19 on the right. So, let's add a coefficient of 9 in front of O₂.
C₉H₂O + 9O₂ → 9CO₂ + H₂O
This is our final balanced equation.
Hope this helps!
Answer:
The answer to your question is letter C.
Explanation:
Reaction
Potassium hydroxide = KOH
Barium chloride = BaCl₂
Potassium chloride = KCl
Barium hydroxide = Ba(OH)₂
KOH + BaCl₂ ⇒ KCl + Ba(OH)₂
Reactant Elements Products
1 K 1
1 Ba 1
2 Cl 1
1 H 2
1 O 2
The reaction is unbalanced
2KOH + BaCl₂ ⇒ 2KCl + Ba(OH)₂
Reactant Elements Products
2 K 2
1 Ba 1
2 Cl 2
2 H 2
2 O 2
Now, the reaction is balanced
Energy is absorbed so, the mass is increased.
energy is released so, the mass is increased
