<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.
Hello!
A) At pH=1
This pH is lower than the value for the pKa, so Acetic acid wouldn't be ionized, but the equilibrium would be displaced to
CH₃COOHCH₃COOH ⇄ CH₃COO⁻ + H₃O⁺ (equilibrium displaced to the
left)
The chemical structure for CH₃COOH is the first one in the attached images.
B) At pH=7
This pH is higher than the value for the pKa, so Acetic acid would be ionized, and the equilibrium would be displaced to
CH₃COO⁻
CH₃COOH ⇄ CH₃COO⁻ + H₃O⁺ (equilibrium displaced to the
right)
The chemical structure for CH₃COO⁻ is the second one in the attached images.
Have a nice day!
Answer:
0.55 atm
Explanation:
First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by

where
m = 1.00 g is the mass of the gas
Mm = 44.0 g/mol is the molar mass of the gas
Substituting,

Now we can find the pressure of the gas by using the ideal gas law:

where
p is the gas pressure
V = 1.00 L is the volume
n = 0.0227 mol is the number of moles
R = 0.082 L/(atm K mol) is the gas constant
T = 25.0 C + 273 = 298 K is the temperature of the gas
Solving the formula for p, we find
