Answer:
See explanation
Explanation:
% optical purity = specific rotation of mixture/specific rotation of pure enantiomer * 100/1
specific rotation of mixture = 23°
specific rotation of pure enantiomer = 61°
Hence;
% optical purity = 23/61 * 100 = 38 %
More abundant enantiomer = 100% - 38 % = 62%
Hence the pure (S) carvone is (-) 62° is the more abundant enantiomer.
Enantiomeric excess = 62 - 50/50 * 100 = 24%
Hence
(R) - carvone = 38 %
(S) - carvone = 62%
The statement of the combined gas law for a fixed amount of gas is,
PV/T = constant
Here, the units of pressure and volume must be consistent and the temperature must be the absolute temperature (Kelvin or Rankine).
0.65 atm is equivalent to 494 mmHg
Using the equation:
(755 x 500) / (27 + 273) = (494 x V) / (-33 + 273)
V = 3396 ml = 3.4 liters
Answer:
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Explanation:
Molecular Equation:
Zn°(s) + Fe(NO₃)₂(aq) => Zn(NO₃)₂(aq) + Fe°(s)
Ionic Equation:
Zn°(s) + Fe⁺²(aq) + 2NO₃⁻(aq) => Zn⁺²(aq) + 2NO₃⁻(aq) + Fe°(s)
Net Ionic Equation: => Drop NO₃⁻ as spectator ion
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Answer:
The partial pressure of SO₃ is 82.0 atm
Explanation:
The equilibrium constant Kp is equal to <em>the equilibrium pressure of the gaseous products raised to the power of their stoichiometric coefficients divided by the equilibrium pressure of the gaseous reactants raised to the power of their stoichiometric coefficients</em>.
For the 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 half-life at this temperature is, 3.28 s
Explanation :
To calculate the half-life for second order the expression will be:
![t_{1/2}=\frac{1}{k\times [A_o]}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5Ctimes%20%5BA_o%5D%7D)
When,
= half-life = ?
= initial concentration = 0.45 M
k = rate constant = 
Now put all the given values in the above formula, we get:


Therefore, the half-life at this temperature is, 3.28 s