Once released, acetylcholine is degraded by extracellular enzymes into what product(s)?A) acetate + choline
B) acetyl CoA + choline
C) methylcholine + acetate
D) choline only
E) acetate only

Answers

Answer 1

Answer:

Final Answer:

Once released, acetylcholine is degraded by extracellular enzymes into acetate + choline. Option A is the correct answer.

Explanation:

Acetylcholine (ACh) is a neurotransmitter responsible for transmitting signals across synapses in the nervous system. Once it is released into the synaptic cleft, acetylcholine is rapidly broken down by the enzyme acetylcholinesterase (AChE). The enzymatic degradation of acetylcholine by AChE results in the hydrolysis of the acetyl group, leading to the formation of acetate and choline.

Acetate is a small molecule that is easily cleared from the synaptic cleft, while choline can be taken up by the presynaptic neuron to be reused for the synthesis of new acetylcholine molecules. This degradation process is essential for terminating the signal transmission and preventing continuous stimulation of the postsynapticneuron.

Option A is the correct answer.

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Answer 2

Answer: It’s acetate + choline

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Please answer immediatelyHow many grams of glucose, C6H12O6, are needed to make 100 mL of a 1.5 M solution?

Answers

THE ANSWER BOI:

27 g

HERE IS DA EXPLANATION FAM:

27 g

M(C6H12O6) = 6*12 + 12*1 + 6*16 = 180 g/mol

100 mL = 0.1 L solution

1.5 M = 1.5 mol/L

1.5 mol/L * 0.1 L = 0.15 mol C6H12O6

0.15 mol * 180 g/1 mol = 27 g C6H12O6

Final answer:

To find the needed grams of glucose for a 1.5 M solution in 100ml, you multiply the molarity by the molecular weight of glucose and the volume of the solution. The calculation is 1.5 mol/l * 180 g/mol * 0.1 l = 27 grams. Therefore, 27 grams of glucose is needed.

Explanation:

The problem in question requires you to make use of the formula M= mass (mol)/ Volume (L). To find the required grams of glucose, C6H12O6, for a 1.5 M solution in 100mL, you would multiply the molarity with the molecular weight of glucose - approximately 180g per mole - and the volume of the solution - 0.1L. Therefore, to solve the problem you would calculate it as: 1.5 mol/l X 180 g/mol x 0.1 l which equals 27g. Hence, you need 27 grams of glucose to make a 100ml, 1.5M solution.

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What is a petroleum by-product that has excellent sealing properties to hold moisture in the skin? a) Talcum powder b) Lanolin c) Astringent d) Petroleum jelly

Answers

Answer: D) Petroleum Jelly

Explanation: Petroleum jelly is also known as white soft paraffin or Vaseline. It's a semi-solid mixture of hydrocarbons that can be used as a lubricant and protective ointment for dry or chapped skin. It can also be used to protect against minor cuts and burns.

Consider the following unbalanced redox reaction:MnO4- (aq) + SbH3 (aq) ---> MnO2 (s) + Sb (s)

What is the oxidizing agent in the reaction?

A.) MnO4-

B.) SbH3

C.) MnO2

D.) Sb

Help please?

Answers

You're oxidising agent will be reduced. In the molecule MnO4-, Mn has and oxidation number of +9 and in MnO2 it is +4, therefore its oxidation number has reduced, therefore Mn is reduced, therefore it is the oxidant/oxidising agent.

What is the chemical composition of ammonium sulfate

Answers

ammonium refers to NH₄⁺ and sulfate refers to SO₄²⁻ . That means that ammonium sulfate is composed of NH₄⁺ and SO₄²⁻ with a formula of (NH₄)₂SO₄.

I hope this helps. Let me know if anything is unclear.

Round off the following measurement to three significant digits: 3255 mL. (a) 325 mL (b) 326 mL (c) 3250 mL (D) 3260 mL (e) none of the above

Answers

The answer is B. If you round it correctly you would get 326 ml and this number happens to have three significant digits.

Final answer:

To round off the measurement 3255 mL to three significant digits, the correct answer is (c) 3250 mL.

Explanation:

To round off a measurement to three significant digits, we look at the digit to the right of the third significant digit. If this digit is 5 or greater, we round up. If it is less than 5, we round down. In the given measurement of 3255 mL, the digit to the right of the third significant digit is 5, so we round up. Therefore, the correct answer is (c) 3250 mL.

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In a nuclear reaction, the energy released is equal to 8.1 x 1016 joules. Calculate the mass lost in this reaction. (1 J = 1 kg m2/s2) and the speed of light = 3.00 × 10^8 m/s ]Answer

7.2 x 1032 kg

2.7 x 108 kg

9.0 x 10-1 kg

7.2 x 10-16 kg