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alexandr1967 [171]
3 years ago
10

Hi, can i have help please?

Chemistry
1 answer:
andreev551 [17]3 years ago
6 0

Answer:

what class?

Explanation:

is there codes for each of these

i think that would be easier to figure out

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8.3 × 106 - trust me, it's actually right. You can use the calculator to see if I'm correct. Punch in <span>8.3 × 106 = 6.6</span>
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a 50.0 mL sample of KCl requires 22.40 mL of 0.0229 M Pb(NO3)2 in order to completely titrate it. What is the Molarity of the KC
garik1379 [7]

Answer:

0.02052M

Explanation:

First, we need to write a balanced equation for the reaction. This is illustrated below:

2KCl + Pb(NO3)2 → 2KNO3 + PbCl2

The following were obtained from the question:

Molarity of Pb(NO3)2 = 0.0229M

Volume of Pb(NO3)2 = 22.40 = 22.4/1000 = 0.0224L

Number of mole of Pb(NO3)2 =?

Recall:

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.0229x0.0224

Mole of Pb(NO3)2 = 5.13x10^-4mole

From the equation,

1mole of Pb(NO3)2 required 2moles KCl.

Therefore, 5.13x10^-4mole of Pb(NO3)2 will require = 5.13x10^-4x2 = 1.026x10^-3mole of KCl.

Now we can use this amount (i.e 1.026x10^-3mole) to find the molarity of KCl. This is illustrated below:

Mole of KCl = 1.026x10^-3mole

Volume of KCl = 500mL = 50/1000 = 0.05L

Molarity =?

Molarity = mole /Volume

Molarity of KCl = 1.026x10^-3/0.05

Molarity of KCl = 0.02052M

8 0
4 years ago
When (R)-6-bromo-2,6-dimethylnonane is dissolved in CH3OH, nucleophilic substitution yields an optically inactive solution. When
3241004551 [841]

Answer:

See explanation

Explanation:

The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.

On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.

7 0
3 years ago
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