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2 years ago

Which of the following would be a reasonable synthesis of CH3CH2CH2CH2OH?Group of answer choices

Chemistry

1 answer:

vampirchik [111]2 years ago

8 0

Answer:

B. 1-Butene rightarrow (1) BH3: THF (2)H202, OH-

Explanation:

In the hydroboration of alkenes, an alkene is hydrated to form an alcohol with anti-Markovnikov orientation.

the reagent BH₃:THF is the way that borane is used in organic reactions. The BH₃ adds to the double bond of an alkene to form an alkyl borane. Peroxide hydrogen in basic medium oxidizes the alkyl borane to form an alcohol. Indeed, hydroboration-oxidation converts alkenes to alcohols by adding water through the double bond, with anti-Markovnikov orientation.

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How is phosphorylation of glyceraldehyde 3-phosphate in the payment phase of glycolysis different from phosphorylation of glucos

svet-max [94.6K]

In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. During this phase, the conversion of glyceraldehyde-3-phosphate to pyruvate and the coupled formation of ATP take place. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the payoff phase occurs twice per molecule of glucose.

Glyceraldehyde 3-phosphate dehydrogenase Simultaneous oxidation and phosphorylation of G3P produce 1,3-bisphosphoglycerate (1,3-BPG) and nicotine adenine dinucleotide (NADH).

The divalent cation also affected the response of the enzyme from the endosperm and shoots to adenine nucleotides and inorganic pyrophosphate.

This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.

For more information on phosphorylation click on the link below:

brainly.com/question/7465103

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7 0

2 months ago

Hi! help me please. CALCULATE THE NUMBER OF MOLECULES FOUND IN 8 MOL OF WATER?

kvasek [131]

Answer:

so the number of molecules in 8 moles of water is 4.8176×10^24

Explanation:

number of moles=8moles

avogadro's number=6.022×10²³

number of molecules=?

as we know that

$number of moles=\frac{number of molecules}{avogadro's number}$

substituting the equation

number of moles × avogdro's number=number of molecules

number of molecules=8moles×6.022×10²³

number of molecules=4.8176×10^24

i hope this will help you :)

4 0

2 years ago

For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

2 years ago

Quiz due soon need help ASAP

Rom4ik [11]

Answer:

Explanation:

Because in product 3CuSO4 is present it means 3 SO4 sulphate ions are present

8 0

1 year ago

An unknown quantity of methane gas, CHA, is held in a 2.00-liter container at 77°C. The pressure inside the container is 3.00 am

kolbaska11 [484]

Answer:

$0.21\text{ mol}$

Explanation:

Here, we want to calculate the number of moles of methane in the container

From the ideal gas law:

$\begin{gathered} PV\text{ = nRT} \\ n\text{ = }\frac{PV}{RT} \end{gathered}$

where:

P is the pressure inside the container which is 3 atm

V is the volume of the container which is 2 L

R is the molar gas constant which is 0.0821 Latm/mol.k

T is the temperature in Kelvin (we convert the temperature in Celsius by adding 273 : 273 + 77 = 350 K)

n is the number of moles that we want to calculate

Substituting the values, we have it that:

$n=\text{ }\frac{3\times2}{0.0821\text{ }\times350}\text{ = 0.21 mol}$

6 0

3 months ago