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vitfil [10]
3 years ago
9

If you find the mass of a sample to be 42.1 g and the actual mass is 42.0 g. What wold be the percent error

Chemistry
1 answer:
Ksivusya [100]3 years ago
4 0

Answer:

0.24%

Explanation:

The following data were obtained from the question:

Measured value = 42.1 g

Actual value = 42 g

Percentage error =?

Percentage error can be obtained by using the following formula:

Percentage error = |Measured value – Actual value | / Actual value × 100

With the above formula, we can obtain the percentage error as follow:

Measured value = 42.1 g

Actual value = 42 g

Percentage error =?

Percentage error = |42.1 – 42| / 42 × 100

Percentage error = 0.1/42 × 100

Percentage error = 10/42

Percentage error = 0.24%

Therefore, the percentage error is 0.24%

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Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.

The product formed on reaction with water would be a 50:50 mixture of

2S-hexane-2-ol. and 2R-hexane-2-ol.

Explanation:

2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good  nucleophile .

The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction  takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.

In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.

When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .

The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.

The SN¹ reaction is a 2 step reaction , in  the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.

In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.

The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.

Kindly refer the attachment for reaction mechanism and structure of products.

8 0
3 years ago
Determine the molecular formula of a compound with an empirical formula of NH2 and a formula mass of 32.06 amu
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N2H4

<span>Each nitrogen weighs 14.01 and each H weighs 1.01. !4.01+14.01+1.01+1.01 = 32.06 (roughly) </span>

4 0
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alex41 [277]

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3 years ago
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dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

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V = 10 mL solution

V_{added} = 12.20 mL

V_{total} = 22.20 mL

then,

M 0.103 M of NaOH,

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V_{total \ base} = 10.38 mL

Now,

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V_{sample} = 10  \ mL

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                 =\frac{ 0.53457 }{10}

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6 0
3 years ago
1. Three different observers measure the melting point of a substance and agree on the
Naddik [55]

d. precise but not accurte

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4 years ago
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