Calculate the volume in liters of a 0.360 mol/l barium acetate solution that contains 100.g of barium acetate (Ba(C2H3O2)2) . Ro
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Answer:
95.6 %
Explanation:
For this question, we will have <u>2 reactions</u>, the formation of the <u>grignard reagent</u> and the <u>formation of the alcohol</u>. The first step then is the calculation of the <u>maximum amount</u> of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the <u>following conversion ratios</u>:
Molar mass of Mg = 24 g/mol
Molar mass of phenylmagnesium bromide ()= 157 g/mol
Density of bromobenzene= 1.5 g/mL
Molar ratio between Mg and = 1:1
The smallest value is the mol of bromobenzene therefore <u>0.0402 mol</u> of phenylmagnesium bromide would be produced.
The next step is repite the same steps for the reaction of <u>formation of the alcohol</u>. Therefore we have to find the moles of methyl benzoate, so:
Molar mass of methyl benzoate: 136.14 g/mol
The we have to <u>divide by the coefficient</u> of each reactive in the balance reaction. So:
Therefore the <u>limiting reagent</u> would be the phenylmagnesium bromide. Now, the <u>molar ratio</u> between the phenylmagnesium bromide and triphenyl carbinol is <u>2:1</u>, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to <u>grams of triphenyl carbinol</u>:
Molar mass of triphenyl carbinol= 260.33 g/mol
Finally, we have to <u>divide</u> the obtanied solid by the calculated one:
<span>Uranium-236 is intermediate nuclei. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-bomb.
Following is the reaction involved in above process:
</span>
+ 
→ 
→ 
+ 
+ 3 <span> + 177 MeV
</span>
Here,
= Fission material,
= projectile,
= intermediate nuclei,
and = Fission product
Following is the reaction involved in above process:
</span>
</span>
Here,