Answer:
The partial pressure of SO₃ is measured at 82.0 atm.
Explanation:
The equilibrium constant Kp is defined as the ratio of the equilibrium pressures of the gaseous products, each raised to the power of their respective coefficients in the reaction, divided by the pressures of the gaseous reactants raised to their coefficients.
For the given reaction,
2 SO₂(g) + O₂(g) → 2 SO₃(g)
![Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm](https://tex.z-dn.net/?f=Kp%20%3D%200.345%20%3D%20%5Cfrac%7B%28pSO_%7B3%7D%29%5E%7B2%7D%20%7D%7B%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%2835.0%29%5E%7B2%7D%20%5Ctimes%2015.9%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%2082.0%20atm)
Answer:
The solution to your inquiry is C = 0.000333 kcal/g°C
or C = 0.333 cal/g°C
Explanation:
Data
Q = 1.67 kcal
mass = 79.2 g
ΔT = 63.3°C
Formula
Q = mCΔT
Solving for C
C = Q/mΔT
Substituting values
C = 1.67/(79.2 x 63.3)
Simplifying
C = 1.67 / 5013.4
Final Result
C = 0.000333 kcal/g°C
or C = 0.333 cal/g°C
<span> </span><span>1. Other (Alcohol)
3. Acidic
5. Salt
</span>
Answer:
No
Explanation:
No. The demonstration in question does not infringe upon the conservation of mass.
The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction; however, mass can change from one form to another during the process.
In this instance, although the remnants of the paper weigh 0.5 g compared to the original weight of 2.5 g, the ashes and gases produced during combustion account for the missing mass of the paper.
The portion that has been burnt has transformed into other states. If the gas and ashes are adequately contained, they will correspond to the weight of the original paper when added to the remaining paper.
Problem 2
You begin with 216 micrograms of Fermium - 253. After three days, the quantity halves, resulting in 108 micrograms left.
Another three days pass. Beginning with 108 micrograms, this amount gets halved again, leaving 54 micrograms.
Finally, after another three-day span, starting from 54 micrograms, you again halve this amount to reach 27 micrograms.
#days Amount in micrograms
0 216
3 108
6 54
9 27
Problem One
Your example is Nitrogen. Begin by completing the table, then formulate some rules to help prepare for possible alternate elements in the test. This approach is quite useful.
Table
Bond Energy Kj/Mol Bond Length pico meters
N - N 167 145
N=N 418 125
N≡N 942 110
Rules
As the number of bonds INCREASES, the energy within the bond also INCREASES
As the number of bonds INCREASES, the distance of the bond DECREASES.
