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snow_tiger [21]
3 years ago
14

Is a straight chain, a linear chain and an unbranched chain the same thing?

Chemistry
2 answers:
Minchanka [31]3 years ago
8 0

Answer:

All carbon atoms can be connected to a continuous chain in the straight-chain alkane. In alkanes of a branched-chain, all carbon atoms cannot be connected to a continuous chain as a branch or side chain is part of the carbon chain.

Explanation:

Schach [20]3 years ago
3 0

Answer:

In the straight-chain alkanes, each carbon is bound to its two neighbors and to two hydrogen atoms. (In red) Exceptions are the two terminal carbon nuclei, which are bound to only one carbon atom and three hydrogen atoms. (In blue)

Explanation:

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Can someone help?? This is really hard.
Sergeu [11.5K]

Answer:

See Explanation

Explanation:

Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.

2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4

2^1 = 2^1

The rate of reaction is first order with respect to Hbn

Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.

1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4

3^1 = 3^1

The reaction is also first order with respect to CO

b) The overall order of reaction is 1 + 1=2

c) The rate equation is;

Rate = k [CO] [Hbn]

d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4  /[5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4/6.7 * 10^-7

k = 4.7 * 10^2 mmol-1 L s-1

e) The reaction occurs in one step because;

1) The rate law agrees with the experimental data.

2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.

8 0
2 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
2 years ago
Resumen de la alquimia farmaceutica
arsen [322]

Answer: English??

Explanation:

8 0
2 years ago
Before Darwin published his ideas,all thoughts and theories on design of species came from which book?
ira [324]

Answer:

Darwin had arrived at a complete theory of evolution by 1839, but it was to be another 20 years before he published his ideas of evolution through natural selection in his epochal book On the Origin of Species by Means of Natural Selection.

5 0
2 years ago
Step five: measure the mass of the water copper tear the balance put calorimeter no lid on the balance measure the mass to the n
OLEGan [10]

Answer:

The picture is the correct answers

Explanation:

5 0
3 years ago
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