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White raven [17]

1 year ago

11

Amobarbital sodium react with ethanolic sodium hyrooxide

1 answer:

CaHeK987 [17]1 year ago

6 0

Amobarbital (like all barbiturates) works by being incontestible to the GABAA receptor at either the alpha or the beta subunit.

<h3>What is the mechanism of amobarbital?</h3>

Amobarbital (like all barbiturates) works by binding to the GABAA receptor at either the alpha or the beta subunit. These are compulsory sites that are distinct from GABA itself and also distinct from the benzodiazepine binding site.

Amobarbital is a barbiturate classified as having a halfway duration of action, meaning that the effects of the drug can last from 4-6 amobarbital increases the effects of benazepril by apparatus: pharmacodynamic synergism.

So we can conclude that Amobarbital, 5-ethyl-5-isoamyl barbituric acid like all barbiturates.

Learn more about Amobarbital here: brainly.com/question/7237163

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The reaction ch4 (g) + 2 o2 (g) ? co2 (g) + 2 h2o (g) is: the reaction ch4 (g) + 2 o2 (g) co2 (g) + 2 h2o (g) is: an exothermic

barxatty [35]

All of the above
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3 years ago

How much cesium (half-life = 2 years) would remain from a 10 g sample after 4 years?

Ksenya-84 [330]

It would be 2.5 g or A.

7 0

2 years ago

Based on the graph below, which statement

Novay_Z [31]

Answer:

4. Atmospheric pressure decreases as altitude increases.

Explanation:

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3 0

2 years ago

As energy from the Sun hits the ocean, that energy transforms the water in

nignag [31]

Answer:

F

Explanation:

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7 0

2 years ago

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can <u>put the values into the equation</u>:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can <u>solve for the initial temperature of gold</u>, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

5 0

3 years ago