Answer:
Explanation:
= Concentration of stock solution
= Concentration of solution
= Volume of stock solution = 19 mL
= Volume of solution = 0.31 L= 310 mL
We have the relation
The concentration of the diluted solution will be 0.613 times the concentration of the stock solution.
Answer:
<u>Bond energy of</u> (A) C≡C > (B) C=C
<u>and, </u>(C) C=N > (D) C-N
Explanation:
Bond energy refers to the amount of energy required to break a bond or the energy released when a bond is formed. Bond energy of a covalent bond suggests the bond strength of the chemical bond and depends on the <u>bond length and bond order of the chemical bond. </u>
<u>The bond energy of a chemical bond increases with the bond order and decreases with the bond length</u>. As, length of a bond decreases with increase in the bond order.
First pair: (A) C≡C (B) C=C
The bond order of C≡C - 3, the bond order of C=C - 2
Since the bond order: C≡C > C=C
Bond length: C≡C < C=C
<u>Therefore, bond energy of (A) C≡C > (B) C=C</u>
Second pair: (C) C=N (D) C-N
The bond order of C=N - 2, the bond order of C-N - 1
Since the bond order: C=N > C-N
Bond length: C=N < C-N
<u>Therefore, bond energy of (C) C=N > (D) C-N</u>
Answer:
Without the tilt, lunar eclipse will happen every month.
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
Learn more about pH of buffer:
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