Answer:
The half-life of a radioisotope describes the amount of time it takes for said isotope to decay to one-half the original amount present in the sample.
Nitrogen-13, because it has a half-life of ten minutes, will experience two half-lives over the course of the twenty minute period. This means that 25% of the isotope will remain after this.
0.25 x 128mg = 32mg
32mg of Nitrogen-13 will remain after 20 minutes.
Answer:
Explanation:
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In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
We plug in the mass of water, temperature change and specific heat to obtain:
Now, this enthalpy of reaction corresponds to the combustion of propyne:
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
Now, we solve for the enthalpy of formation of C3H4 as shown below:
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
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Answer:
<h2>
= (
1.08 /
2.2
) 100% = 49%</h2>
Explanation:
Balanced Equation: 2CH₃CH₃(g) + 5O₂(g) → 2CO₂(g) + 6H₂O(g)
Calculate moles of CH₃CH₃ and O₂
1.2 ₃₃ (
1 ₃₃/
30.0694 ₃₃
) = 0.040 ₃₃
8.6 ₂ (
1 2/
31.998 ₂
) = 0.27 ₃₃
Find limiting reagent 0.040 ₃₃ (
5 ₂/
2 ₃₃
) = 0.10 ₂
CH₃CH₃ is the limiting Reagent
CH₃CH₃ (L.R.) O₂ CO₂ H₂O
Initial (mol) 0.040 0.27 0 0
Change
(mol)
-2x=-0 -5x=
-0.10 +2x=+0.040 +6x=+0.12
Final (mol) 0 0.117 0.040 0.12
0.040 − 2 = 0 = 0.020
Determine percent yield
0.12 ₂ (
18.0148 ₂
/1 ₂
) = 2.2 ₂
= (
1.08 /
2.2
) 100% = 49%