Answer:
see explanation
Explanation:
The process of ionization to produce cations is endothermic. For formation of Ca⁺² two ionization steps need be illustrated as follows...
1st ionization step: Ca° + 590Kj => Ca⁺ + e⁻
2nd ionization step: Ca⁺ + 1151Kj => Ca⁺² + e⁻
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Net Ionization Rxn: Ca° + 1741Kj => Ca⁺² + 2e⁻
<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:
First
divide each element by its Molecular Mass to get their respective moles
Then Divide through by the lowest of the moles
You'll have the ratio of Carbon Hydrogen and Oxygen to be
C2H3O
Given Molecular Mass=184.27
C2H3On=184.27
n(12x2 + 1x3 + 16) =184.27
Evaluating this... You'll have n=4.3
Pls check if you assigned the correct value to each element
Answer:-
Water is highly ordered. In water each oxygen atom is connected to others around it through hydrogen bonding via bridging hydrogen atoms. When a salt like NaCl is dissolved, some of these Hydrogen bonds break.
When a salt like NaCl dissolves in water, the NaCl breaks in to ions Na+ and Cl-.
The water molecules now surround these ions.
The slightly negative oxygen end of water molecule gets near the Na+, while the slightly positive Hydrogen of water molecule gets near the Cl-.
So before salt sample dissolve, the water molecules were highly ordered due to hydrogen bonding. Now after salt dissolve there is a decrease in order and thus an increase in disorder of the water molecules.
Due to increase in disorder, entropy which is a measure of disorder increases. Since entropy increases, delta S for the process is positive.
