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1 year ago

Ibuprofen can be found in 800 mg doses in over-the-counter analgesics, such as Advil and Motrin. How many grams of iburofen

Chemistry

1 answer:

geniusboy [140]1 year ago

7 0

0.8g Ibuprofen

1 g = 10^-3g = .001g

Ibuprofen has 800 mg doses in over-the-counter analgesic

800g = 800 × .001

= 0.8g

Ibuprofen is Nondteriodal Anti-inflammatory Drug (NSAID)
Ibuprofen's Mechanism of Action is Decreases inflammation, pain, and fever through inhibition of cyclooxygenase activity and prostaglandin synthesis
nonsteroidal anti-inflammatory medication (NSAID) used for pain relief and to reduce fever by stops inflammation and by blocking formation of cyclo-oxygenase (COX-2) a chemical mediator of inflammatory chemicals. i.e prostaglandins
It comes under the Class analgesic (reduce pain) and antipyretic (FIRE - reduce fever)
e side effects of ibuprofen NSAID are peripheral edema, fluid retention with edema, tinnitus, purpura, petechiae, anorexia, diarrhea, rash, nausea, vomiting, fatigue, dizziness, lightheadedness, anxiety, confusion

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3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

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Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

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3 years ago