Answer:
a. 7278 K
b. 4.542 × 10⁻³¹
Explanation:
a.
Let´s consider the following reaction.
N₂(g) + O₂(g) ⇄ 2 NO(g)
The reaction is spontaneous when:
ΔG° < 0 [1]
Let's consider a second relation:
ΔG° = ΔH° - T × ΔS° [2]
Combining [1] and [2],
ΔH° - T × ΔS° < 0
ΔH° < T × ΔS°
T > ΔH°/ΔS°
T > (180.5 × 10³ J/mol)/(24.80 J/mol.K)
T > 7278 K
b.
First, we will calculate ΔG° at 25°C + 273.15 = 298 K
ΔG° = ΔH° - T × ΔS°
ΔG° = 180.5 kJ/mol - 298 K × 24.80 × 10⁻³ kJ/mol.K
ΔG° = 173.1 kJ/mol
We can calculate the equilibrium constant using the following expression.
ΔG° = - R × T × lnK
lnK = - ΔG° / R × T
lnK = - 173.1 × 10³ J/mol / (8.314 J/mol.K) × 298 K
K = 4.542 × 10⁻³¹
Answer:
Percent error = -22%
Explanation:
Given data:
Density of aluminium = 2.1,2.12 and 2.09 g/cm³
Accepted value of density of aluminium = 2.7 g.cm³
Percent error = ?
Solution:
First of all we will calculate the average of data measured by students.
Average value = 2.1 g/cm³ + 2.12 g/cm³+ 2.09 g/cm³ / 3
Average value = 6.31 g/cm³/ 3
Average value = 2.1 g/cm³
Now we will calculate the percent error.
Percent error = [ Measured value - accepted value / accepted value ]× 100
Percent error = [ 2.1 g/cm³ - 2.70 g/cm³ / 2.70 g/cm³ ]× 100
Percent error = [-0.6/2.70 g/cm³ ] × 100
Percent error = -0.22 × 100
Percent error = -22%
Answer: hello some part of your question is missing below is the missing part
In an experiment to determine the % of ascorbic acid in Vitamin C Tablets by Titration with Potassium Bromate,
answer:
Oxidation half reaction of Vitamin C
Explanation:
The solution will turn cloudy dark purple even after reaching endpoint when allowed to settle with time. because of the Oxidation half reaction of Vitamin C. also during the Titration process few drops of starch solution will be added to help determine the endpoint of the experiment .
Density = Mass divided by Volume
So what you do is 80.00 ÷ 0.99755.
Which I believe would equal <span>80.1964813794
Hope this helps! </span>
it shoudl be : 2Fe(NO3)3 + 3MgSO4 → 3Mg(NO3)2 + Fe2(SO4)3
(the difference is Fe2(SO4)3 has no coefficient)
