Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.
Explanation:
1) Molarity of 0.250 L HCl solution : 0.0328 M
Moles of HCl in 0.250 L solution = 0.0082 moles
2) Molarity of 0.100 L NaOH solution : 0.0245 M
Moles of NaOH in 0.100 L solution = 0.00245 moles
3) Concentration of hydrochloric acid in the resulting solution.
0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.
Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L
Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles = 0.00575 moles
Molarity of HCl left un-neutralized :
0.0164 molar concentration of hydrochloric acid in the resulting solution.
<em>Answer:</em>
The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M
<em>Data Given:</em>
Ksp of CaSO4 = 2.4 × 10∧-5
CaSO4 ⇔ Ca+2 + SO4∧-2
<em>Solution:</em>
Ksp = [Ca+2].[ SO4∧-2]
2.4 × 10∧-5 = [x].[x]= x²
x = 4.9×10∧-3 M
<em>Result:</em>
- The conc. of Ca+2 ion is 4.9×10∧-3 M
Answer:
The change in color.
Explanation:
The apple turn brown in color because of the oxidation process. When the oxygen and water molecules in air react with it, oxidation take place. The oxidation process is very efficient in ambient temperature.
For example, if the peal off apple is placed into the refrigerator it take a time to got oxidize and turn brown, but if it is placed in room temperature it quickly turn brown.
when oxygen is react with peel off apple , it trigger the polyphenol oxidase enzyme to oxidize the phenolic compound and quinones are formed which then react with amino acids and produced brown color.
Electrons fill orbitals in order of increasing energy from left to right. As the group number increase also the number of valence electorns of each group will increases
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
