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Norma-Jean [14]
4 years ago
10

In a classroom which comparison would a teacher most likely use for describing a mole

Chemistry
1 answer:
irina [24]4 years ago
3 0
Mole - one of the most important concepts in chemistry - is a kind of link to go from the microworld of atoms and molecules in a normal macrocosm grams and kilograms.
In chemistry often have to consider large numbers of atoms and molecules. For fast and efficient calculation made using the weighing method. But it is necessary to know the weight of individual atoms and molecules. In order to identify the molecular weight must be added the weight of all atoms in the compound.
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3 years ago
How many grams of NaI are needed to prepare 400.0 mL of 0.0500 M NaI solution?
Simora [160]

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:

\frac{400mL}{1000mL}x0.0500moles=0.0200moles

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:

150\frac{g}{mol}x0.0200moles = 3g

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3 years ago
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Amphetamine (C9H13N)(C9H13N) is a weak base with a pKbpKb of 4.2. You may want to reference (Pages 710 - 713) Section 16.8 while
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Answer:

pH = 10.38

Explanation:

  • C9H13N ↔ C9H20O3N+  +  OH-

∴ 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

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