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Usimov [2.4K]
3 years ago
14

Provide an appropriate alkyne starting material A and intermediate product B. Omit byproducts. The number of carbon atoms in the

starting material should be the same as in the final product.
Chemistry
1 answer:
nadezda [96]3 years ago
6 0

Answer:

C4H6 + 2HBr = C4H8Br2

Explanation:

Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of CnH2n−2. They are unsaturated hydrocarbons.

Furthermore, Halogenation reaction of any alkyne such as but-2-yne with any of the hydro halogenated compounds such as HBr, HCl, HF or HI will give an alkane that will have the same number of carbon atoms in the starting material and in the final product.

Halogenation reaction is a reaction that occurs when one or more halogens are added to a substance. Halogens is the seventh column in the periodic table and it include fluorine, chlorine, bromine, iodine, and astatine.

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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
8 months ago
Lucia is playing with magnetic toy vehicles. She has two identical toy vehicles (purple and pink) that start on opposite sides o
pentagon [3]

Answer:

The potential energy of both toy vehicles (purple and pink) decreased. Since the pink toy was moved closer to the magnet, it will have less potential energy because of the short distance it will take to travel to the magnet. Although the purple toy is now closer to the magnet, it is still pretty far and will have a somewhat big potential energy when traveling to the magnet.

Explanation:

Hey, I'm in middle school and I had the same question for a science test, I'm not sure if I am correct but this is what I have.

3 0
2 years ago
Read 2 more answers
Can you tell me name of Chemicals in food?​
Artyom0805 [142]

Answer:

Guar gum

sodium nitrite

artificial food colorings

monosodium glutamate

etc

6 0
2 years ago
A chemist must dilute 97.1 ml of aqueous magnesium fluoride solution until the concentration falls to 389 microMolarity . He'll
mojhsa [17]

Answer:

0.302L

Explanation:

<em>...97.1mL of 1.21m M aqueous magnesium fluoride solution</em>

<em />

In this problem the chemist is disolving a solution from 1.21mM = 1.21x10⁻³M, to 389μM = 389x10⁻⁶M. That means the solution must be diluted:

1.21x10⁻³M / 389x10⁻⁶M = 3.11 times

As the initial volume of the original concentration is 97.1mL, the final volume must be:

97.1mL * 3.11 = 302.0mL =

0.302L

6 0
2 years ago
Acids will corrode most _____.<br> bases<br> liquids<br> metals<br> gases
aliya0001 [1]
Acids corrode most metals
3 0
3 years ago
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