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Irina18 [472]
2 years ago
13

Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 ∘C? (HINT: the act

ivation energy for this reaction is 160 kJ/mol) T(K) 1/T(K−1) ln k 462.9 2.160×10−3 -10.589 472.1 2.118×10−3 -9.855 503.5 1.986×10−3 -7.370 524.4 1.907×10−3 -5.757 Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 ? (HINT: the activation energy for this reaction is 160 ) ln 462.9 -10.589 472.1 -9.855 503.5 -7.370 524.4 -5.757 8.1×10−15 s−1 2.2×10−13 s−1 2.7×10−9 s−1 2.0×10−1 s−1 9.2×103 s−1

Chemistry
1 answer:
svp [43]2 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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Nickel-63 has a half life of 92 hours. If a
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Answer:

6.25 g of nickel -63 will be left

Explanation:

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2 years ago
The electron dot structure for CI is
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The correct option is C.
A Lewis dot diagram is a representation of the valence electron of  an atom, which uses dot around the symbol of the atom. Chlorine has seven electrons in its outermost shell, these seven electrons are arranged in form of dot around the atom of chlorine. If you count the number of dot given in option C, you will notice that they are seven.
6 0
2 years ago
Read 2 more answers
The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &S carvone is measured a
shusha [124]

Answer:

See explanation

Explanation:

% optical purity = specific rotation of mixture/specific rotation of pure enantiomer  * 100/1

specific rotation of mixture = 23°

specific rotation of pure enantiomer = 61°

Hence;

% optical purity = 23/61 * 100 = 38 %

More abundant enantiomer = 100% - 38 % = 62%

Hence the pure  (S) carvone is (-) 62° is the more abundant enantiomer.

Enantiomeric excess = 62 - 50/50 * 100 = 24%

Hence

(R) - carvone  =  38 %

(S) - carvone = 62%

7 0
1 year ago
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(f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of d and 0.005 mole of l? (assume a 1-dm path
never [62]
<span>Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.</span>
4 0
2 years ago
Read 2 more answers
An unknown compound melting at 131 - 133 C. It is thought to be one of the following compounds: trans-cinnamic acid (133-134); b
Vesnalui [34]

Answer:

benzamide

Explanation:

Compound            melting Point ,ºC          Melting Pont Mixture, ºC

       X                          131 - 133

trans-cinnamic            133 - 134                      110 - 120

acid

benzamide                 128 - 130                       130-132

malic acid                   131   -133                        114 -124

Benzoin                      135 - 137                        108 - 116

The compound X is benzamide since the melting point range is the one closest to this compound (  130-132 ºC)

The reason there is not an exact match is not due due to the presence of impurities. The presence of impurities always lower the melting point ( it is a coligative property such as the melting point depresion of salt and water )

The reason for the deviation must be  be some other factors such as preparation of the sample in the capillary, errors in reading the thermometer, rate of heating, etc.

5 0
2 years ago
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