balance & tell the types of equations:
_____ p + _____o₂ → _____p₂o₅
_____ na + _____ h₂o → _____ naoh + _____h₂
_____ ag₂o → _____ ag + _____o₂
_____ co₂ + _____ h₂o → _____ c₆h₁₂o₆ + _____o₂
_____ nabr + _____ caf₂ → _____ naf + _____ cabr₂

Question

Accepted Answer

### First Equation: $\boldsymbol{P + O_2
ightarrow P_2O_5}$
# Explanation:
## Step1: Balance O atoms
On right, $P_2O_5$ has 5 O. Left has $O_2$. Find LCM of 2 and 5, which is 10. So put 5 in front of $O_2$ (5×2=10 O) and 2 in front of $P_2O_5$ (2×5=10 O).
Now equation: $P + 5O_2
ightarrow 2P_2O_5$

## Step2: Balance P atoms
Right has 2×2 = 4 P. So put 4 in front of P.
Balanced: $4P + 5O_2
ightarrow 2P_2O_5$

## Step3: Identify reaction type
Two substances (P, $O_2$) combine to form one ($P_2O_5$) → **Synthesis Reaction**

### Second Equation: $\boldsymbol{Na + H_2O
ightarrow NaOH + H_2}$
# Explanation:
## Step1: Balance H atoms
Right: $NaOH$ has 1 H, $H_2$ has 2 → total 3? Wait, better: Left $H_2O$ has 2 H per molecule. Let's balance O first (only in $H_2O$ and $NaOH$). $NaOH$ has 1 O, $H_2O$ has 1 O. So balance Na and O together. Let's try putting 2 in front of $Na$, $H_2O$, $NaOH$.
Now: $2Na + 2H_2O
ightarrow 2NaOH + H_2$

## Step2: Check H
Left: 2×2 = 4 H. Right: 2×1 (from NaOH) + 2 (from $H_2$) = 4 H. Balanced.

## Step3: Reaction type
$Na$ displaces H from $H_2O$ → **Single Displacement (Metal + Water → Base + $H_2$)**

### Third Equation: $\boldsymbol{Ag_2O
ightarrow Ag + O_2}$
# Explanation:
## Step1: Balance O atoms
Left: $Ag_2O$ has 1 O. Right: $O_2$ has 2. Put 2 in front of $Ag_2O$ (2 O) and 1 in front of $O_2$ (2 O? Wait, 2×1=2 O, 1×2=2 O. Now: $2Ag_2O
ightarrow Ag + O_2$

## Step2: Balance Ag atoms
Left: 2×2 = 4 Ag. So put 4 in front of Ag.
Balanced: $2Ag_2O
ightarrow 4Ag + O_2$

## Step3: Reaction type
One compound breaks into two (Ag and $O_2$) → **Decomposition Reaction**

### Fourth Equation: $\boldsymbol{CO_2 + H_2O
ightarrow C_6H_{12}O_6 + O_2}$ (Photosynthesis)
# Explanation:
## Step1: Balance C atoms
Right: $C_6H_{12}O_6$ has 6 C. So put 6 in front of $CO_2$.
Now: $6CO_2 + H_2O
ightarrow C_6H_{12}O_6 + O_2$

## Step2: Balance H atoms
Right: 12 H. Left: $H_2O$ has 2 H per molecule. So put 6 in front of $H_2O$ (6×2=12 H).
Now: $6CO_2 + 6H_2O
ightarrow C_6H_{12}O_6 + O_2$

## Step3: Balance O atoms
Left: 6×2 (CO₂) + 6×1 (H₂O) = 12 + 6 = 18 O.
Right: $C_6H_{12}O_6$ has 6 O, $O_2$ has 2x O. So 6 + 2x = 18 → 2x=12 → x=6. Put 6 in front of $O_2$.
Balanced: $6CO_2 + 6H_2O
ightarrow C_6H_{12}O_6 + 6O_2$

## Step4: Reaction type
$CO_2$ and $H_2O$ react, with $O_2$ as product (reverse of combustion) → **Synthesis (or Photosynthesis, a type of redox)**

### Fifth Equation: $\boldsymbol{NaBr + CaF_2
ightarrow NaF + CaBr_2}$
# Explanation:
## Step1: Balance Br atoms
Right: $CaBr_2$ has 2 Br. Left: $NaBr$ has 1 Br. So put 2 in front of $NaBr$.
Now: $2NaBr + CaF_2
ightarrow NaF + CaBr_2$

## Step2: Balance Na and F atoms
Left: 2 Na, 2 F (from $CaF_2$? Wait, $CaF_2$ has 2 F. Right: $NaF$ has 1 Na and 1 F. So put 2 in front of $NaF$ (2 Na, 2 F).
Balanced: $2NaBr + CaF_2
ightarrow 2NaF + CaBr_2$

## Step3: Reaction type
Ions (Na⁺, Br⁻, Ca²⁺, F⁻) exchange partners → **Double Displacement Reaction**

# Answer:
1. $4P + 5O_2
ightarrow 2P_2O_5$ (Synthesis)
2. $2Na + 2H_2O
ightarrow 2NaOH + H_2$ (Single Displacement)
3. $2Ag_2O
ightarrow 4Ag + O_2$ (Decomposition)
4. $6CO_2 + 6H_2O
ightarrow C_6H_{12}O_6 + 6O_2$ (Synthesis/Photosynthesis)
5. $2NaBr + CaF_2
ightarrow 2NaF + CaBr_2$ (Double Displacement)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

First Equation: $\boldsymbol{P + O_2

Step1: Balance O atoms

Step2: Balance P atoms

Step3: Identify reaction type

Second Equation: $\boldsymbol{Na + H_2O

Step1: Balance H atoms

Step2: Check H

Step3: Reaction type

Third Equation: $\boldsymbol{Ag_2O

Step1: Balance O atoms

Step2: Balance Ag atoms

Step3: Reaction type

Fourth Equation: $\boldsymbol{CO_2 + H_2O

Answer: