Answer:
The answer to your question is letter A. 65.9%
Explanation:
Data
actual yield = 1.20 g
theoretical yield = 1.82 g
Formula
percentage yield = 
Substitution
percentage yield = 
Simplification and result
percentage yield = 0.6593 x 100
percentage yield = 65.9 %
Answer:
After 5 second 25% C-15 will remain.
Explanation:
Given data:
Half life of C-15 = 2.5 sec
Original amount = 100%
Sample remain after 5 sec = ?
Solution:
Number of half lives = T elapsed / half life
Number of half lives = 5 sec / 2.5 sec
Number of half lives = 2
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
Thus after 5 second 25% C-15 will remain.
The balanced reaction would be:<span>
C12H22O11 + 12O2 = 12CO2 + 11H2O
We are given the amount of oxygen used in the combustion. This will be the starting point of our calculation. We use the ideal gas equation to find for the number of moles.
n = PV / RT = 1.00(250 L) / (0.08206 atm L/mol K ) 273 K
n= 11.16 mol O2
</span>11.16 mol O2<span> (12 mol CO2 / 12 mol O2) = 11.16 mol CO2
V = nRT/P =</span>11.16 mol CO2<span> x 273 K x 0.08206 atm L/mol K / 1 atm
V=250 L</span>
Physics<span> beyond the Standard </span>Model<span> (BSM) refers to the theoretical developments needed to ... Yet, the Standard </span>Model<span> does not supply any fundamental </span>particles<span> that are good dark matter candidates. .... compared to their mass—while formation of structures </span>similar<span> to the galaxies in our </span>universe<span>requires cold dark matter.</span>