<h3>1</h3>
Species shown in bold are precipitates.
- Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
- Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃
- Cu(NO₃)₂ + 2 KI → CuI₂ + 2 KI
- Cu(NO₃)₂ + 2 KOH → Cu(OH)₂ + 2 KNO₃
- Cu(NO₃)₂ + Na₂C₂O₄ → CuC₂O₄ + 2 NaNO₃
- Ni(NO₃)₂ + 2 KOH → Ni(OH)₂ + 2 KNO₃
- Ni(NO₃)₂ + Na₂C₂O₄ → NiC₂O₄ + 2 NaNO₃
- Zn(NO₃)₂ + 2 KOH → Zn(OH)₂ + 2 KNO₃
- Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃
<h3>2</h3>
A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.
Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.
<h3>3</h3>
Compare the first and last row:
Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.
As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.
<h3>4</h3>
Compare the second and third row:
Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.
Answer:
Ethanol + Potassium Permanganate + Sulfuric Acid = Acetic Acid + Manganese(II) Sulfate + Water + Potassium Sulfate
Explanation:
Answer:
2B2 + 3O2 → 2B2O3
Explanation:
Balance The Equation: B2 + O2 = B2O3
1. Label Each Compound With a Variable
aB2 + bO2 = cB2O3
2. Create a System of Equations, One Per Element
B: 2a + 0b = 2c
O: 0a + 2b = 3c
3. Solve For All Variables (using substitution, gauss elimination, or a calculator)
a = 2
b = 3
c = 2
4. Substitute Coefficients and Verify Result
2B2 + 3O2 = 2B2O3
L R
B: 4 4 ✔️
O: 6 6 ✔️
hope this helps!
False everything involves matter
