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hydrogen peroxide can be used to disinfect and clean surfaces around th…

Question

hydrogen peroxide can be used to disinfect and clean surfaces around the home. it breaks down over time into oxygen gas and water. consider the following reaction: some amount of hydrogen peroxide (h₂o₂) breaks down to produce 10 molecules of oxygen (o₂) and 20 molecules of water (h₂o). complete the table below. chemical element | number of atoms in the reaction h | o | during this reaction, how many molecules of hydrogen peroxide (h₂o₂) react?

Explanation:

Part 1: Completing the table for number of atoms (H and O)
For Hydrogen (H):

Step1: Find H from $H_2O$

Each $H_2O$ has 2 H atoms. There are 20 $H_2O$ molecules. So H from $H_2O$: $2\times20 = 40$.
(No H in $O_2$ or reactant $H_2O_2$ considered here for product side, but since reaction is balanced, reactant $H_2O_2$ will supply H. But for counting in reaction (product side, and reactant side should match), so total H atoms: 40.

For Oxygen (O):

Step1: Find O from $O_2$

Each $O_2$ has 2 O atoms. 10 $O_2$ molecules: $2\times10 = 20$.

Step2: Find O from $H_2O$

Each $H_2O$ has 1 O atom. 20 $H_2O$ molecules: $1\times20 = 20$.

Step3: Total O atoms

Sum O from $O_2$ and $H_2O$: $20 + 20 = 40$.

Part 2: Number of $H_2O_2$ molecules reacting

The balanced reaction for decomposition of $H_2O_2$ is: $2H_2O_2
ightarrow 2H_2O + O_2$ (but let's derive from products).
From products: 20 $H_2O$ (each has 2 H, so 40 H) and 10 $O_2$ (20 O) + 20 O from $H_2O$ = 40 O.
Each $H_2O_2$ has 2 H and 2 O. Let number of $H_2O_2$ be $x$.
For H: $2x = 40$ (from earlier, total H is 40) $\Rightarrow x = 20$.
Check O: $2x = 40$ (total O is 40) $\Rightarrow x = 20$. Also, from the product $O_2$ (10 molecules, 20 O) and $H_2O$ (20 O), total O 40. Each $H_2O_2$ has 2 O, so $2x = 40 \Rightarrow x = 20$. Also, from the reaction stoichiometry: if 20 $H_2O$ and 10 $O_2$ are produced, the reaction is $2H_2O_2
ightarrow 2H_2O + O_2$ scaled? Wait, no: Let's see moles (molecules) ratio.
From $H_2O$: 20 molecules. From $O_2$: 10 molecules. The balanced equation is $2H_2O_2
ightarrow 2H_2O + O_2$. So for 2 $H_2O$ and 1 $O_2$, we need 2 $H_2O_2$. Here, $H_2O$ is 20 (which is 10 times 2) and $O_2$ is 10 (10 times 1). So $H_2O_2$ should be 10 times 2 = 20.

Answer:

s:

  • H atoms: 40
  • O atoms: 40
  • $H_2O_2$ molecules: 20

(For the table: H: 40, O: 40; and $H_2O_2$: 20)