The standard state formation reaction is a chemical reaction in which one moles of substance in its standard state is formed from its constituent element in their standard state.All the substance must be in their most stable state at 100kpa and 25 degrees celsius.
therefore for HF is
1/2H2 +1/2F2 =HF
Answer:
Explanation:
From the correct question above:
The reaction can be represented as:
From the above reaction; the ICE table can be represented as:
I (mol/L) 0.086 0.28 0 0
C -4x -3x +2x +6x
E 0.086 - 4x 0.28 - 3x +2x +6x
At equilibrium;
The water vapor = 
![\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }](https://tex.z-dn.net/?f=%5Ctext%7Bequilibrium%20constant%7D%20%20%28%7Bk_c%7D%29%20%3D%20%20%5Cdfrac%7B%20%5BN_2%5D%5E2%20%5BH_2O%5D%5E6%20%7D%7B%20%5B%5BNH_3%5D%5E4%5D%20%5BO_2%5D%5E3%20%7D)
Replacing the value of x, we have:
c. a tertiary alcohol; when a ketone reacts with a grignard reagent followed by protonation a tertiary alcohol is formed.
More about tertiary alcohol:
No hydrogen atoms are bonded to the functional group's carbon in a tertiary alcohol. Alcohols that have a hydroxyl group bonded to the carbon atom and are linked to three alkyl groups are referred to as tertiary alcohols. These alcohols' structural makeup largely determines their physical characteristics.
This -OH group's existence enables alcohols to create hydrogen bonds with the atoms next to them. Because of this weak connection, alcohols have higher boiling points than their alkane counterparts.
The alcohol is referred to as a tertiary (3°) alcohol if the carbon atom carrying the alcohol group is connected to three other carbon atoms in the alcohol molecule.
Learn more about tertiary alcohol here:
brainly.com/question/17419424
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Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
