Answer:
A. 0.143 M
B. 0.0523 M
Explanation:
A.
Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).
KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄
The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:
1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol
The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.
5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:
M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M
B.
Let's consider the neutralization of potassium hydroxide and perchloric acid.
KOH + HClO₄ → KClO₄ + H₂O
When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.
Answer : The standard cell potential of the reaction is, -1.46 V
Explanation :
The given balanced cell reaction is,
Here, chromium (Cr) undergoes oxidation by loss of electrons and act as an anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
The standard values of cell potentials are:
Standard reduction potential of lead ![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
Standard reduction potential of chromium ![E^0_{[Cr^{3+}/Cr]}=1.33V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D%3D1.33V)
Now we have to calculate the standard cell potential for the following reaction.
![E^0=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)
Therefore, the standard cell potential of the reaction is, -1.46 V
Fill in the blanks in order w/ these numbers:
1. 2 1 2 1
2. 3 1 1
3. 1 2 1 2
4. 2 1 1
5. 4 3 2
6. 1 1 2
Hope this helps.
Molarity is defined as the number of moles of solute per 1 L of solvent.
the mass of NaCl in the solution is 87.75 g
number of moles of NaCl is calculated by dividing mass present by molar mass
number of NaCl moles = 87.75 g / 58.44 g/mol = 1.502 mol
the number of NaCl moles in 500 mL is - 1.502 mol
therefore number of NaCl moles in 1000 mL is - 1.502 mol/ 500 mL x 1000 mL/L = 3.004 mol
molarity of NaCl is - 3.004 M
answer is D. 3.00 M
