use the codes below to determine the type of reaction
s=synthesis d=decomposition sr=single replacement dr=double replacement c=combustion
1. $ \ce{p + o2 - p4o} $
2. $ \ce{c5h9o + o2 - co2 + h2o} $
3. $ \ce{hgo - hg + o2} $
4. $ \ce{al2o3 - al + o2} $
5. $ \ce{cl2 + nabr - nacl + br2} $
6. $ \ce{h2 + n2 - nh3} $
7. $ \ce{na + br2 - nabr} $
8. $ \ce{cucl2 + h2s - cus + hcl} $
9. $ \ce{hgo + cl2 - hgcl + o2} $
10. $ \ce{c + h2 - ch4} $
11. $ \ce{k(clo3) - kcl + o2} $
12. $ \ce{s8 + f2 - sf6} $
13. $ \ce{mg + o2 - mgo} $
14. $ \ce{c8h16 + o2 - co2 + h2o} $
15. $ \ce{bacl2 + na2(so4) - nacl + ba(so4)} $
16. $ \ce{nh3 + h2(so4) - (nh4)2(so4)} $
17. $ \ce{pb + h3(po4) - h2 + pb3(po4)2} $
18. $ \ce{hbr + al(oh)3 - h2o + albr3} $
19. $ \ce{mgcl2 + li2(co3) - mg(co3) + licl} $
complete the reactant or product in the following reactions. indicate type of reaction.

Question

use the codes below to determine the type of reaction
s=synthesis  d=decomposition  sr=single replacement  dr=double replacement  c=combustion
1. $ \ce{p + o2 -> p4o} $
2. $ \ce{c5h9o + o2 -> co2 + h2o} $
3. $ \ce{hgo -> hg + o2} $
4. $ \ce{al2o3 -> al + o2} $
5. $ \ce{cl2 + nabr -> nacl + br2} $
6. $ \ce{h2 + n2 -> nh3} $
7. $ \ce{na + br2 -> nabr} $
8. $ \ce{cucl2 + h2s -> cus + hcl} $
9. $ \ce{hgo + cl2 -> hgcl + o2} $
10. $ \ce{c + h2 -> ch4} $
11. $ \ce{k(clo3) -> kcl + o2} $
12. $ \ce{s8 + f2 -> sf6} $
13. $ \ce{mg + o2 -> mgo} $
14. $ \ce{c8h16 + o2 -> co2 + h2o} $
15. $ \ce{bacl2 + na2(so4) -> nacl + ba(so4)} $
16. $ \ce{nh3 + h2(so4) -> (nh4)2(so4)} $
17. $ \ce{pb + h3(po4) -> h2 + pb3(po4)2} $
18. $ \ce{hbr + al(oh)3 -> h2o + albr3} $
19. $ \ce{mgcl2 + li2(co3) -> mg(co3) + licl} $
complete the reactant or product in the following reactions. indicate type of reaction.

Accepted Answer

Let's solve problem 1: $ \text{P} + \text{O}_2
ightarrow \text{P}_4\text{O}_{10} $ (first, balance the equation) and determine the reaction type.

### Explanation:
#### Step 1: Balance the chemical equation
- Reactants: P (1 atom), $ \text{O}_2 $ (2 O atoms)
- Products: $ \text{P}_4\text{O}_{10} $ (4 P atoms, 10 O atoms)
- Balance P: Multiply P by 4: $ 4\text{P} + \text{O}_2
ightarrow \text{P}_4\text{O}_{10} $
- Balance O: $ \text{O}_2 $ has 2 O, $ \text{P}_4\text{O}_{10} $ has 10 O. Multiply $ \text{O}_2 $ by 5: $ 4\text{P} + 5\text{O}_2
ightarrow \text{P}_4\text{O}_{10} $

#### Step 2: Determine reaction type
- Synthesis reaction (S) is when two or more substances combine to form a single product. Here, P and $ \text{O}_2 $ combine to form $ \text{P}_4\text{O}_{10} $, so it's a synthesis reaction.

### Answer:
Balanced equation: $ 4\text{P} + 5\text{O}_2
ightarrow \text{P}_4\text{O}_{10} $, Reaction type: S (Synthesis)

---

Let's solve problem 3: $ \text{HgO}
ightarrow \text{Hg} + \text{O}_2 $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation
- Reactants: $ \text{HgO} $ (1 Hg, 1 O)
- Products: Hg (1 Hg), $ \text{O}_2 $ (2 O)
- Balance O: Multiply $ \text{HgO} $ by 2: $ 2\text{HgO}
ightarrow \text{Hg} + \text{O}_2 $
- Balance Hg: Multiply Hg by 2: $ 2\text{HgO}
ightarrow 2\text{Hg} + \text{O}_2 $

#### Step 2: Determine reaction type
- Decomposition reaction (D) is when a single compound breaks down into two or more simpler substances. Here, $ \text{HgO} $ decomposes into Hg and $ \text{O}_2 $, so it's a decomposition reaction.

### Answer:
Balanced equation: $ 2\text{HgO}
ightarrow 2\text{Hg} + \text{O}_2 $, Reaction type: D (Decomposition)

---

Let's solve problem 5: $ \text{Cl}_2 + \text{NaBr}
ightarrow \text{NaCl} + \text{Br}_2 $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation
- Reactants: $ \text{Cl}_2 $ (2 Cl), $ \text{NaBr} $ (1 Na, 1 Br)
- Products: $ \text{NaCl} $ (1 Na, 1 Cl), $ \text{Br}_2 $ (2 Br)
- Balance Br: Multiply $ \text{NaBr} $ by 2: $ \text{Cl}_2 + 2\text{NaBr}
ightarrow \text{NaCl} + \text{Br}_2 $
- Balance Na and Cl: Multiply $ \text{NaCl} $ by 2: $ \text{Cl}_2 + 2\text{NaBr}
ightarrow 2\text{NaCl} + \text{Br}_2 $

#### Step 2: Determine reaction type
- Single Replacement (SR) reaction: A more reactive element replaces a less reactive element in a compound. Cl (more reactive than Br) replaces Br in $ \text{NaBr} $, so it's a single replacement reaction.

### Answer:
Balanced equation: $ \text{Cl}_2 + 2\text{NaBr}
ightarrow 2\text{NaCl} + \text{Br}_2 $, Reaction type: SR (Single Replacement)

---

Let's solve problem 7: $ \text{Na} + \text{Br}_2
ightarrow \text{NaBr} $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation
- Reactants: Na (1 Na), $ \text{Br}_2 $ (2 Br)
- Products: $ \text{NaBr} $ (1 Na, 1 Br)
- Balance Br: Multiply $ \text{NaBr} $ by 2: $ \text{Na} + \text{Br}_2
ightarrow 2\text{NaBr} $
- Balance Na: Multiply Na by 2: $ 2\text{Na} + \text{Br}_2
ightarrow 2\text{NaBr} $

#### Step 2: Determine reaction type
- Synthesis reaction (S): Two substances (Na and $ \text{Br}_2 $) combine to form a single product ($ \text{NaBr} $), so it's a synthesis reaction.

### Answer:
Balanced equation: $ 2\text{Na} + \text{Br}_2
ightarrow 2\text{NaBr} $, Reaction type: S (Synthesis)

---

Let's solve problem 9: $ \text{HgO} + \text{Cl}_2
ightarrow \text{HgCl} + \text{O}_2 $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation (Note: Correct product should be $ \text{HgCl}_2 $ maybe a typo, assuming $ \text{HgCl}_2 $)
- Let's correct the product to $ \text{HgCl}_2 $ for a valid reaction.
- Reactants: $ \text{HgO} $ (1 Hg, 1 O), $ \text{Cl}_2 $ (2 Cl)
- Products: $ \text{HgCl}_2 $ (1 Hg, 2 Cl), $ \text{O}_2 $ (2 O)
- Balance O: Multiply $ \text{HgO} $ by 2: $ 2\text{HgO} + \text{Cl}_2
ightarrow \text{HgCl}_2 + \text{O}_2 $
- Balance Hg and Cl: Multiply $ \text{HgCl}_2 $ by 2: $ 2\text{HgO} + 2\text{Cl}_2
ightarrow 2\text{HgCl}_2 + \text{O}_2 $

#### Step 2: Determine reaction type
- Single Replacement (SR): Cl (more reactive than O in $ \text{HgO} $) replaces O? Wait, no, maybe a double replacement? Wait, no, $ \text{HgO} $ and $ \text{Cl}_2 $: Cl replaces O? Alternatively, if it's $ \text{HgO} $ decomposing with Cl, but likely a single replacement. However, the original problem has $ \text{HgCl} $ which is incorrect (should be $ \text{HgCl}_2 $). Assuming correction, it's a single replacement or maybe a redox reaction. But based on the code, maybe the intended reaction is decomposition or single replacement. However, with the given product $ \text{HgCl} $ (which is unstable), it's better to check the original problem again. Alternatively, maybe it's a typo and the product is $ \text{HgCl}_2 $. For the sake of solving, let's proceed with $ \text{HgCl}_2 $.

#### Step 2 (Revised):
- If we assume the reaction is $ 2\text{HgO} + 2\text{Cl}_2
ightarrow 2\text{HgCl}_2 + \text{O}_2 $, it's a single replacement (Cl replaces O in $ \text{HgO} $) or a redox reaction. But based on the code, maybe the type is Single Replacement (SR) or Decomposition? No, decomposition is one compound breaking down. Here, two reactants, so maybe Single Replacement.

### Answer:
Balanced equation (corrected product): $ 2\text{HgO} + 2\text{Cl}_2
ightarrow 2\text{HgCl}_2 + \text{O}_2 $, Reaction type: SR (Single Replacement) (assuming product correction)

---

Let's solve problem 11: $ \text{K(ClO}_3\text{)}
ightarrow \text{KCl} + \text{O}_2 $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation
- Reactants: $ \text{KClO}_3 $ (1 K, 1 Cl, 3 O)
- Products: KCl (1 K, 1 Cl), $ \text{O}_2 $ (2 O)
- Balance O: Multiply $ \text{KClO}_3 $ by 2: $ 2\text{KClO}_3
ightarrow \text{KCl} + \text{O}_2 $
- Balance K and Cl: Multiply KCl by 2: $ 2\text{KClO}_3
ightarrow 2\text{KCl} + 3\text{O}_2 $

#### Step 2: Determine reaction type
- Decomposition reaction (D): A single compound ($ \text{KClO}_3 $) breaks down into two simpler substances (KCl and $ \text{O}_2 $), so it's a decomposition reaction.

### Answer:
Balanced equation: $ 2\text{KClO}_3
ightarrow 2\text{KCl} + 3\text{O}_2 $, Reaction type: D (Decomposition)

---

Let's solve problem 13: $ \text{Mg} + \text{O}_2
ightarrow \text{MgO} $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation
- Reactants: Mg (1 Mg), $ \text{O}_2 $ (2 O)
- Products: $ \text{MgO} $ (1 Mg, 1 O)
- Balance O: Multiply $ \text{MgO} $ by 2: $ \text{Mg} + \text{O}_2
ightarrow 2\text{MgO} $
- Balance Mg: Multiply Mg by 2: $ 2\text{Mg} + \text{O}_2
ightarrow 2\text{MgO} $

#### Step 2: Determine reaction type
- Synthesis reaction (S): Two substances (Mg and $ \text{O}_2 $) combine to form a single product ($ \text{MgO} $), so it's a synthesis reaction.

### Answer:
Balanced equation: $ 2\text{Mg} + \text{O}_2
ightarrow 2\text{MgO} $, Reaction type: S (Synthesis)

---

Let's solve problem 15: $ \text{BaCl}_2 + \text{Na}_2\text{SO}_4
ightarrow \text{NaCl} + \text{BaSO}_4 $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation
- Reactants: $ \text{BaCl}_2 $ (1 Ba, 2 Cl), $ \text{Na}_2\text{SO}_4 $ (2 Na, 1 S, 4 O)
- Products: $ \text{NaCl} $ (1 Na, 1 Cl), $ \text{BaSO}_4 $ (1 Ba, 1 S, 4 O)
- Balance Na and Cl: Multiply $ \text{NaCl} $ by 2: $ \text{BaCl}_2 + \text{Na}_2\text{SO}_4
ightarrow 2\text{NaCl} + \text{BaSO}_4 $

#### Step 2: Determine reaction type
- Double Replacement (DR): Ions of two compounds exchange places. $ \text{Ba}^{2+} $ and $ \text{Na}^+ $ exchange with $ \text{Cl}^- $ and $ \text{SO}_4^{2-} $, so it's a double replacement reaction.

### Answer:
Balanced equation: $ \text{BaCl}_2 + \text{Na}_2\text{SO}_4
ightarrow 2\text{NaCl} + \text{BaSO}_4 $, Reaction type: DR (Double Replacement)

---

Let's solve problem 17: $ \text{Pb} + \text{H}_3\text{PO}_4
ightarrow \text{H}_2 + \text{Pb}_3(\text{PO}_4)_2 $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation
- Reactants: Pb (1 Pb), $ \text{H}_3\text{PO}_4 $ (3 H, 1 P, 4 O)
- Products: $ \text{H}_2 $ (2 H), $ \text{Pb}_3(\text{PO}_4)_2 $ (3 Pb, 2 P, 8 O)
- Balance Pb: Multiply Pb by 3: $ 3\text{Pb} + \text{H}_3\text{PO}_4
ightarrow \text{H}_2 + \text{Pb}_3(\text{PO}_4)_2 $
- Balance P: Multiply $ \text{H}_3\text{PO}_4 $ by 2: $ 3\text{Pb} + 2\text{H}_3\text{PO}_4
ightarrow \text{H}_2 + \text{Pb}_3(\text{PO}_4)_2 $
- Balance H: Multiply $ \text{H}_2 $ by 3: $ 3\text{Pb} + 2\text{H}_3\text{PO}_4
ightarrow 3\text{H}_2 + \text{Pb}_3(\text{PO}_4)_2 $

#### Step 2: Determine reaction type
- Single Replacement (SR): Pb (a metal) replaces H in $ \text{H}_3\text{PO}_4 $ (an acid), so it's a single replacement reaction.

### Answer:
Balanced equation: $ 3\text{Pb} + 2\text{H}_3\text{PO}_4
ightarrow 3\text{H}_2 + \text{Pb}_3(\text{PO}_4)_2 $, Reaction type: SR (Single Replacement)

---

Let's solve problem 19: $ \text{MgCl}_2 + \text{Li}_2\text{CO}_3
ightarrow \text{MgCO}_3 + \text{LiCl} $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation
- Reactants: $ \text{MgCl}_2 $ (1 Mg, 2 Cl), $ \text{Li}_2\text{CO}_3 $ (2 Li, 1 C, 3 O)
- Products: $ \text{MgCO}_3 $ (1 Mg, 1 C, 3 O), $ \text{LiCl} $ (1 Li, 1 Cl)
- Balance Li and Cl: Multiply $ \text{LiCl} $ by 2: $ \text{MgCl}_2 + \text{Li}_2\text{CO}_3
ightarrow \text{MgCO}_3 + 2\text{LiCl} $

#### Step 2: Determine reaction type
- Double Replacement (DR): Ions of two compounds exchange places. $ \text{Mg}^{2+} $ and $ \text{Li}^+ $ exchange with $ \text{Cl}^- $ and $ \text{CO}_3^{2-} $, so it's a double replacement reaction.

### Answer:
Balanced equation: $ \text{MgCl}_2 + \text{Li}_2\text{CO}_3
ightarrow \text{MgCO}_3 + 2\text{LiCl} $, Reaction type: DR (Double Replacement)

---

Let's solve problem 2: $ \text{C}_5\text{H}_9\text{O} + \text{O}_2
ightarrow \text{CO}_2 + \text{H}_2\text{O} $ (balance and determine type)

### Explanation:
#### Step 1: Balance the chemical equation (Combustion reaction, C)
- Reactants: $ \text{C}_5\text{H}_9\text{O} $ (5 C, 9 H, 1 O), $ \text{O}_2 $ (2 O)
- Products: $ \text{CO}_2 $ (1 C, 2 O), $ \text{H}_2\text{O} $ (2 H, 1 O)
- Balance C: Multiply $ \text{CO}_2 $ by 5: $ \text{C}_5\text{H}_9\text{O} + \text{O}_2
ightarrow 5\text{CO}_2 + \text{H}_2\text{O} $
- Balance H: Multiply $ \text{H}_2\text{O} $ by $ \frac{9}{2} $ (or 9 and adjust O): $ \text{C}_5\text{H}_9\text{O} + \text{O}_2
ightarrow 5\text{CO}_2 + \frac{9}{2}\text{H}_2\text{O} $
- Multiply all by 2 to eliminate fractions: $ 2\text{C}_5\text{H}_9\text{O} + 13\text{O}_2
ightarrow 10\text{CO}_2 + 9\text{H}_2\text{O} $

#### Step 2: Determine reaction type
- Combustion (C): A

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step 1: Balance the chemical equation

Step 2: Determine reaction type

Step 1: Balance the chemical equation

Step 2: Determine reaction type

Step 1: Balance the chemical equation

Step 2: Determine reaction type

Answer: