Answer:
Exergonic ,Endergonic,low concentration area,high
Explanation:
In exergonic reaction,certain molecules are broken down;in the process they release energy which is captured when high energy molecules(such as ATP and NADH) are formed.
The breakdown of these molecules can be coupled to thermodynamically unfavorable processes such as Endergonic reactions or pumping og hydrogen ion from low concentration areas to high concentration areas.
These types of molecules are called hydrates. They have a certain number of moles attached to the salt. Their characteristic is being hygroscopic. That means that when they are exposed to air, they readily solvate.
The formula for Manganese Fluoride Decahydrate will involve the formula Mn, F and H₂O. In ionic form, Manganese is Mn⁺² while fluoride is in F⁻. When they are brought together, their superscripts are 'cross-multiplied' and becomes their respective subscripts. The compound becomes MnF₂. Then, we add the decahydrate which means 10 moles of H₂O. Hence, the formula for Manganese Fluoride Decahydrate is MnF₂*10H₂O.
Answer:
a) C6H5COOH + H2O ↔ H3O+ + C6H5COO-
b) [ H3O+ ] = 2.517 E-3 M
c) pH = 2.599
Explanation:
a) balanced equation:
C6H5COOH + H2O ↔ H3O+ + C6H5COO-
⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5
mass balance:
0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)
charge balance:
[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant
⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)
b) (2) in (1):
⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]
⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]
⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5
⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0
⇒ [ H3O+ ] = 2.517 E-3 M
c) pH = - log [ H3O+ ]
⇒ pH = - Log ( 2.517 E-3 )
⇒ pH = 2.599
Explanation:
We are given: entropy of Fe2O3 = 90J/K.mol
: entropy of C = 5.7J/K.mol
: entropy of Fe = 27.2J/K.mol
: entropy of CO = 198J/K.mol
Answer:
The correct answer is C.
I think convergent but could be wrong
