<u>Answer:</u> The given chemical reaction can be classified as synthesis and exothermic.
<u>Explanation:</u>
A synthesis reaction is defined as the reaction where two small chemical species combine in their elemental state to form a single large chemical species.
Exothermic reactions are defined as the reactions in which heat is released by the reaction. The heat is written on the product side of the reaction.
For the given chemical reaction:
The above chemical reaction is a type of synthesis and exothermic as two substances in their elemental state are combining. Also, heat is getting released in the reaction.
Hence, the given chemical reaction can be classified as synthesis and exothermic.
Silver (Ag) is likely to be less reactive
The lower you go, the more acidic. The higher you go, the more alkaline. Your answer would most likely be 6.
Original molarity was 1.7 moles of NaCl
Final molarity was 0.36 moles of NaCl
Given Information:
Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity
Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity
1. Solve for the molarity of the original (concentrated) solution.
Molarity (M) = moles of solute (mol) / liters of solution (L)
Convert the given information to the appropriate units before plugging in and solving for molarity.
Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)
2. Solve for the molarity of the final (diluted) solution.
Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.
Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution
Molarity (M) of the final solution = 0.36 M NaCl
I hope this helped:))
<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.
