Answer:
C2H4O2 + 2O2 —> 2CO2 + 2H20
The coefficients are 1, 2, 2, 2
Explanation:
C2H4O2 + O2 —> CO2 + H20
To balance the above, do the following
Since there are 2 carbon on the left, put 2 in front of CO2, we have
C2H4O2 + O2 —> 2CO2 + H20
We have 4 hydrogen on the left, to balance it put 2 in front of H2O i.e
C2H4O2 + O2 —> 2CO2 + 2H20
Now, we have a total of 6 oxygen on the right side. To balance it, put 2 in front of O2:
C2H4O2 + 2O2 —> 2CO2 + 2H20
The coefficients are 1, 2, 2, 2
Based on nuclear stability, the symbol for the most likely product nuclide obtained when nitrogen-13 undergoes decay is coming from the following equation:
₇N¹³ → Positron₊₁⁰ + ₆C¹³
So the correct answer will be ₆C¹³
Largest
K
Ca
Ga
Ge
As
Br
Kr
Smallest
Answer:
Explanation:
Remark
First of all you have to identify what is happening. You make the following observations.
The solution in the test tube in a cold water bath is colorless.
As you add heat, the color changes from colorless to brown.
What you are told
N2O4 is colorless
NO2 is brownish red.
What you conclude
The reaction is endothermic. That means it requires heat to happen. An endothermic reaction is
A + Heat ===> B
So you have three possible correct answers
N2O4 + 14 kCal ===> 2NO2
N2O4 ====> 2NO2 dH = 14 kCal
N2O4 ====> 2NO2 - 14 kCal
I can't read the last 4 equations.
Other answers
As the temperature increased, the N2O4 became less.
As the temperature increased, the products were favored. (color change)
The reaction is exothermic (gives off heat) when the reaction goes from
NO2 - Heat ===> N2O2
Answer:
ΔG° = 1747.523
Explanation:
The parameters mentioned are;
Gibbs Free energy ΔG°
Equilibrium constant Kc
Temperature T = 37 + 273 = 310 (upon conversion to kelvin temperature)
The formular relating all three parameters is given as;
ΔG° = -RTlnKc
Where; R = rate constant = 8.314 J⋅K−1⋅mol−1
Upon solving;
ΔG° = - 8.314 * 310 * ln(1.97)
ΔG° = 1747.523