4. This neurotransmitter is a part of the brain's "reward system"

Chemistry

1 answer:

Sliva [168]2 years ago

8 0

Answer:

Dopamine

Explanation:

The regions of the brain comprising the “reward system” use the neurotransmitter dopamine to communicate.

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A + B = AB

Ahat [919]

The greatest amount of AB would be produced if the equilibrium constant of the reaction is equal to $1.0 \;X \;10^5$ . Hence, option D is correct.

<h3>What is an equilibrium constant?</h3>

A number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

The equilibrium constant expression is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants.

If the value of K is greater than 1, the products in the reaction are favoured. If the value of K is less than 1, the reactants in the reaction are favoured.

Hence, option D is correct.

Learn more about the equilibrium constant here:

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7 0

1 year ago

Mashcka [7]

Answer:

It is 34

Explanation:

4 0

2 years ago

The ka of hf is 6.8 x 10-4. what is the ph of a 0.35 m solution of hf?

damaskus [11]

When the reaction equation is:

HF ↔ H+ + F-

and when the Ka expression
= concentration of products/concentration of reactions

so, Ka = [H+][F-]/[HF]

when we assume:

[H+] = [F-] = X

and [HF] = 0.35 - X

So, by substitution:

6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X

∴ X = 0.015 M

∴[H+] = X = 0.015

when PH = -㏒[H+]

∴PH = -㏒0.015

= 1.8

6 0

2 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .