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stellarik [79]
3 years ago
7

HELP ASAP

Chemistry
2 answers:
Romashka [77]3 years ago
6 0

Answer:

1) 206 g /mol

2) 0.971 mmol or 9.71 X 10⁻⁴ mol

3) 7.768 mmol or 7.768 X 10⁻³ mol

Explanation:

1) Let us calculate the molar mass by adding the atomic mass of each atom present in the given ibuprofen molecule.

Molecular formula : C₁₃H₁₈O₂

Molar mass = 13 X atomic mass of C + 18X atomic mass of H +2 X atomic mass of O

Molar mass = 13 X 12 + 18X1 + 2X16 = 206g/mol

2) moles is calculated as:

moles=\frac{mass}{molarmass}=\frac{200mg}{206}=0.971mmol

This is the moles in each ibuprofen tablet.

3) Now in each dose there are two tablets.

The adult took four doses

The number of tablets taken = 4X2 = 8

the number of moles taken = 8X0.971 mmol = 7.768 mmol

Virty [35]3 years ago
5 0
 <span>To find the molar mass, look at a periodic table for each element. 
Ibuprofen, C13 H18 and O2. Carbon has a molar mass of 12.01 g, Hydrogen has 1.008 g per mole, and Oxygen is 16.00 g per mole. 

C: 13 * 12.01 
H: 18 * 1.008 
O: 2 * 16.00 
Calculate that, add them all together, and that is the molar mass of C13H18O2. 
Molar mass: 206.274 

Next, you have 200mg in each tablet, with a ratio of C13H18O2 (molar mass) in GRAMS per Mole 
So, you need to convert miligrams into grams, which is 200 divided by 1000. 
0.2 g / Unknown mole = 206.274 g / 1 Mole 

This is a cross multiplying ratio where you're going to solve for the unknown moles of grams per tablet compared to the moles per ibuprofen. 
So, it's set up as: 
0.2 g * 1 mole = 206.274 * x 
0.2 = 206.274x 
divide each side by 206.274 to get X alone 
X = 0.00097 
or 9.7 * 10^-4 moles 

The last problem should be easy to figure out now that you have the numbers. 1 dose is 2 tablets, which is the moles we just calculated above, times four for the dosage. 
</span>
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According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

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S+H_2+2O_2\rightarrow H_2SO_4    \Delta H_{formation}=?

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Now adding all the equations, we get the expression for enthalpy of formation of H_2SO_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

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\Delta H=-812.4kJ/mol

Therefore, the enthalpy of formation of H_2SO_4 is, -812.4 kJ/mole

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