In the sequence of nitrogenous bases
Answer:
CHO2- ion
Explanation:
We have the lewis structure of a formate-ion here
This is CHO2-.
The carbon atom is the central atom in the structure. It's the least electronegative atom (C). Carbon has 4 valence electrons. Oxygen has 6 valence electrons.
The carbon will bind with 1 hydrogen atom, this will form 1 single bond, because hydrogen has 1 valence electron.
The carbon will bind with oxygen via a double bond.
Since carbon has only 4 valence electrons, it can only form 1 bond with the other oxygen atom.
There will formed 1 double bond between C and O and 1 single bond between C and O resulting in a negative charged O-atom.
This means there are two resonance structures. for the CHO2- ion
To decrease the rate of the forward reaction you can do the following:
- decrease the concentration of the reactants
- increase the concentration of the products
- increase the pressure so the hydrogen gas will not be easily removed from the reaction
- decrease the temperature, number of collisions will decrease so the rate of the reaction will decrease
The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:
ΔS = nCln(T₂/T₁)
n = number of moles
C = molar heat capacity
T₂ = final temperature
T₁ = initial temperature
We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:
[(3 moles)(273.15 K) + (1 mole)(373.15 K)]/(4 moles) = 298.15 K
Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:
ΔS = (3 moles)(75.3 J/Kmol)ln(298.15/273.15)
ΔS = 19.8 J/K
ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
ΔS = -16.9 J/K
Now we combine the entropy change of each portion of water to get the total entropy change for the system:
ΔS = 19.8 J/K + (-16.9 J/K)
ΔS = 2.9 J/K
The entropy change for combining the two temperatures of water is 2.9 J/K.
