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irinina [24]
3 years ago
7

A formula for a cough syrup contains 60 mg of codeine per fluid ounce. How many mg are contained in one teaspoonful?

Chemistry
1 answer:
Kruka [31]3 years ago
7 0

Answer : The amount of codeine in one teaspoonful is, 30 mg

Explanation : Given,

Amount of codeine per fluid ounce = 60 mg

Now we have to determine the amount of codeine present in one teaspoonful.

As we know that:

1 teaspoonful = 0.5 fluid ounce

As, the amount of codeine per fluid ounce = 60 mg

So, the amount of codeine 0.5 fluid ounce = \frac{0.5}{1}\times 60mg=30mg

Thus, the amount of codeine in one teaspoonful is, 30 mg

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Why should scientists be able to replicate an investigation?
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Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system,
givi [52]

Answer: The order from the Most energy released to most Energy   Absorbed Is given as  2---> 4--->,3-->---> 1 

B)-61.9 kJ

Explanation:

The change in the internal energy of a system  is positive if the reaction absorbs energy and  negative if the reaction releases energy. For a system to cause an increase in volume, it must have very high energy built up to be released.

1. Surroundings get colder and the system decreases in volume. Here, the surrounding absorbs energy  resulting in positive  ΔE

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3. Surroundings get hotter and the system decreases in volume. Although there is a decreased volume, the system is negative because it releases energy

4. Surroundings get hotter and the system does not change in volume.  System is negative because it releases energy even thgoygh there is no change in volume

Therefore the order from the Most energy released to most Energy   Absorbed Is given as  2---> 4--->,3-->---> 1

b) Using  

 ΔE = q+ w     from 1st law of thermodynamics

 ΔE=  ΔH - P  ΔV

gIven  

 ΔH = -75.0KJ

volume=  A change  from 5.0L TO 2.0L = Final volume - initial volume = 2-5= -3.00L

P= 43.0atm

ΔE=  ΔH - P  ΔV

P  ΔV  = 43 atm x -3 = -129L.atm

We first convert  L-atm to Joules.

1 L-atm = 101.325 Joules.  

129L.atm = 129 x 101.325 = - 13071 J

to KJ becomes

13071/1000 = - 13.071KJ

Recall ΔE=  ΔH - P  ΔV and putting values

ΔE  = -75.0 - (-13.071 KJ)= -75.0 kJ + 13.071 kJ = -61.9 kJ

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