Answer:
-2092 kJ
Explanation:
Let's consider the chemical reaction that causes chromium to corrode in air.
4 Cr + 3 O₂ → 2 Cr₂O₃
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ΔH° - T × ΔS°
where,
- ΔH°: standard enthalpy of the reaction
- ΔS°: standard entropy of the reaction
ΔG° = -2256 kJ - 298 K × (-0.5491 kJ/K)
ΔG° = -2092 kJ
Answer:
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
For two situations and phases, the equation becomes:
Given:
= 13.95 torr
= 144.78 torr
= 25°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
= 298.15 K
= 75°C = 348.15 K
So,
Rydberg Eqn is given as:
1/λ = R [1/n1^2 - 1/n2^2]
<span>Where λ is the wavelength of the light; 2626 nm = 2.626×10^-6 m </span>
<span>R is the Rydberg constant: R = 1.09737×10^7 m-1 </span>
<span>From Brackett series n1 = 4 </span>
<span>Hence 1/(2.626×10^-6 ) = 1.09737× 10^7 [1/4^2 – 1/n2^2] </span>
<span>Some rearranging and collecting up terms: </span>
<span>1 = (2.626×10^-6)×(1.09737× 10^7)[1/16 -1/n2^2] </span>
<span>1= 28.82[1/16 – 1/n2^2] </span>
<span>28.82/n^2 = 1.8011 – 1 = 0.8011 </span>
<span>n^2 = 28.82/0.8011 = 35.98 </span>
<span>n = √(35.98) = 6</span>
Acetaminophen as a chemical formula of C8H9NO2. The molar
masses are:
C8H9NO2 = 151.163 g/mol
C = 12 g/mol
H = 1 g/mol
N = 14 g/mol
O = 16 g/mol
<span>TO get the mass percent, simply multiply the molar mass of
each elements with the number of the
element divide by the molar mass of acetaminophen, that is:</span>
%C = [(12 * 8) / 151.163] * 100% = 63.50%
%H = [(1 * 9) / 151.163] * 100% = 5.954%
%N = [(14 * 1) / 151.163] * 100% = 9.262%
<span>%O = [(16 * 2) / 151.163] * 100% = 21.17% </span>
Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log(
) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log(
) and so 10 = or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
