Answer:
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Answer:
3g of NaI is needed to prepare 400.0 mL of 0.0500 M NaI solution.
Explanation:
A 0.0500M solution means that is 0.0500 moles of NaI per every 1000 mL of solution. Thus, to prepare 400mL:

Also, the molar mass of NaI is 127g/mol + 23g/mol = 150g/mol
Consequently, multiplying the molar mass of NaI by the moles the necessary mass is obtained:

Answer:
pH = 10.38
Explanation:
∴ molar mass C9H13N = 135.21 g/mol
∴ pKb = - log Kb = 4.2
⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]
∴ <em>C</em> sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M
mass balance:
⇒ <em>C</em> sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)
charge balance:
⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water
⇒ [C9H20O3N+] = [OH-].......(2)
(2) in (1):
⇒ [C9H13N] = 1.516 E-3 - [OH-]
replacing in Kb:
⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])
⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0
⇒ [OH-] = 2.3985 E-4 M
∴ pOH = - Log [OH-]
⇒ pOH = 3.62
⇒ pH = 14 - pOH = 14 - 3.62 = 10.38
Idk.... my grade in the school C (about chemistry)
It wasn’t as powerful because your sense of smell enhances your sense of taste