Answer:
Kkkkkkkkkkkkkkk=√π67444
Explanation:
answer=727273737364644464646463636366363+929292929292929339939393
<span>83.9%
First, determine the molar masses of Al(C6H5)3 and C6H6. Start by looking up the atomic weights of the involved elements.
Atomic weight aluminum = 26.981539
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 1.00794 = 258.293239 g/mol
Molar mass C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol
Now determine how many moles of C6H6 was produced
Moles C6H6 = 0.951 g / 78.11184 g/mol = 0.012174851 mol
Looking at the balanced equation, it indicates that 1 mole of Al(C6H5)3 is required for every 3 moles of C6H6 produced. So given the number of moles of C6H6 you have, determine the number of moles of Al(C6H5)3 that was required.
0.012174851 mol / 3 = 0.004058284 mol
Then multiply by the molar mass to get the number of grams that was originally present.
0.004058284 mol * 258.293239 g/mol = 1.048227218 g
Finally, the weight percent is simply the mass of the reactant divided by the total mass of the sample. So
1.048227218 g / 1.25 g = 0.838581775 = 83.8581775%
And of course, round to 3 significant digits, giving 83.9%</span>
Answer:
2.05mg Fe/ g sample
Explanation:
In all chemical extractions you lose analyte. Recovery standards are a way to know how many analyte you lose.
In the problem you recover 3.5mg Fe / 1.0101g sample: <em>3.465mg Fe / g sample. </em>As real concentration of the standard is 4.0 mg / g of sample the percent of recovery extraction is:
3.465 / 4×100 = <em>86,6%</em>
As the recovery of your sample was 1.7mg Fe / 0.9582g, the Iron present in your sample is:
1.7mg Fe / 0.9582g sample× (100/86.6) = <em>2.05mg Fe / g sample</em>
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I hope it helps!
Answer:
The solution to the question is as follows
(a) The rate of ammonia formation = 0.061 M/s
(b) the rate of N₂ consumption = 0.0303 M/s
Explanation:
(a) To solve the question we note that the reaction consists of one mole of N₂ combining with three moles of H₂ to form 2 moles of NH₃
N₂(g) + 3H₂(g) → 2NH₃(g)
The rate of reaction of molecular hydrogen = 0.091 M/s, hence we have
3 moles of H₂ reacts to form 2 moles of NH₃, therefore
0.091 M of H₂ will react to form 2/3 × 0.091 M or 0.061 M of NH₃
Hence the rate of ammonia formation is 0.061 M/s
(b) From the reaction equation we have 3 moles of H₂ and one mole of N₂ being consumed at the same time hence
0.091 M of H₂ is consumed simultaneously with 1/3 × 0.091 M or 0.0303 M of N₂
Therefore the rate of consumption of N₂ = 0.0303 M/s
