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madam [21]
2 years ago
11

A nursing instructor is teaching about pharmacological treatments for attention-deficit/hyperactivity disorder (ADHD). Which inf

ormation about atomoxetine (Strattera) should be included in the lesson plan?
Chemistry
1 answer:
pashok25 [27]2 years ago
5 0

Answer:

The possible numerous

side effects of atomoxetine.

Explanation:

The side effects of atomoxetine are numerous and so it should be included in the lesson plan.

Side effects like Stomach upset, nausea, vomiting, constipation, tiredness, loss of appetite/weight loss, dry mouth, dizziness, drowsiness, trouble sleeping, or decrease in sexual ability/desire may occur.

Also,increased blood pressure, difficulty urinating, fainting,liver damage, abdominal pain, heart attack, stroke.

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liq [111]

Answer:

When cinnamic acid react with bromine ,addition reaction rapidly occur on alkene functional group to form dibromo product

Explanation:

Phenyl ring is an aromatic hydrocarbon ,when aromatic hydrocarbons react with Cl2,Br2 or KMnO4 no reaction occur ,where as unsaturated hydrocarbon  like alkene react .Aromatic hydrocarbon with these reagents undenr different conditions undergoes subtituition reaction.They react with bromine in presence of lewis acid catalyst ferric bromide.

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3 years ago
CH3-CH2-CH=CH-CH=CH2
o-na [289]

Answer:

Hydration (of an alkene)

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8 0
2 years ago
How is nuclear energy generated?
balandron [24]
Nuclear power plants heat water to produce steam. The steam is used to spin large turbines that generate electricity. ... In nuclear fission, atoms are split apart to form smaller atoms, releasing energy. Fission takes place inside the reactor of a nuclear power plant.
3 0
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Alina [70]

Answer:

G

Explanation:

8 0
2 years ago
The temperature of 15.71 grams of gold rises from 32°C to 1,064°C, and then the gold melts completely. If gold’s specific heat i
antoniya [11.8K]

Answer:Total energy gained by 15.71 g is 3090.6471 joules

Explanation:

Given:

Q = heat gained by the 15.71 gram mass of gold

Q=mc\Delta T +m\Delta H_{fusion}

\Delta T=(T_{final}-T_{initial})=(1064^oC-32^oC)=1032^oC

c = specific heat capacity of gold = 0.1291joules/gram^oC

m = mass of gold =15.71 g

Q=15.71 g\times 0.1291 joules/gram^oC (1032^oC) +15.71 g\times 63.5 Joules/gram

Heat gained by gold = 2093.0621 Joules + 997.585 Joules

=3090.6471 joules

6 0
2 years ago
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