Approximate Solution After Shifting Starting Concentration
substituting known compounds into the formula and then approximating the answer by moving around the variables, then getting the percent by multiplying by a hundred. DONT FORGET TO RAISE THE EXPRESSION BY THE NUMBER OF MOLES.
Calculation of an Equilibrium Constant
general method of creating an ICE table. we can move this around accordingly, but make sure that i include all the correct numbers, i could also solve for any number that can be easily given.
ΔSsys=−qrevTsys
this is losing heat, since the heat in reverse and does not go back to it's initial state that would mean it would be the same going forward and that would mean the system will have more entropy to give to the surrounding. so we add the minus here to indicate that this is just going to lose heat.
ΔSuniv=ΔSsys+ΔSsurr
the change in entropy of the universe is equal to the entropy inside of the system plus the entropy of the surrounding of the system.
The spontaneous process whereby the gas contained initially in one flask expands to fill both flasks equally therefore yields an increase in entropy for the system.
the process is described in a way that as the gas expanded to both containers, the level of energy used by the gas to do that was none. So that must mean that space that is being increased is the space that is occupied by the gas in however many systems. Is the meaning of entropy (S).
ΩcΩa
the final divided by the initial, how we get the final microstate is by exponentially multiplying the systems by the number of however many particles are in the system.
ΔS=Sf−Si
change in entropy is equal to the entropy at the end minus the entropy at the beginning. that can be calculated by the Boltzmann constant x the natural log of the final microstate divided by the initial microstate. (yes this can be mathematically be expanded and compressed)
ΔS=qrevT
the entropy is equal to the heat we can give back to the system to reverse it's reaction divided by the total Temp.
From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects.
no energy is being lost in the process of transferring the heat between both the systems in thermodynamics.
−qX>0
less than zero so it's colder.
qX<0
more than zero so it's hotter
3.7 × 10–5 M
try putting this into a ICE table
hydronium
H3O
pOH and the pH of a 0.0125-M solution
easier to start with one expression then plug it into pKw
log[H3O+]=−7.3
bring the log to the other side of the expression, which would be the base ten on the other side.
hydronium ion
pH
Calculate the pH of the solution at 25 °C.
look at what they are asking us, from there use the pH or pOH expression and just solve from there.
10−7
if the exponent is equal to -7 then try testing it for pH and pOH.
80 °C
the temp going up brings the pH constant down from 14 to whatever lower and it's the same way inversed as well.
hydronium
H3O
hydroxide
OH
negative sign indicates that the reaction is exothermic
remember to just show at the end if the reaction is endo or exo.
Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = − heat taken in by water
rearrange the equation to show what's losing heat and what is gaining heat here.
q is given a negative (-) sign when the system releases heat to the surroundings (exothermic); q is given a positive (+) sign when the system absorbs heat from the surroundings (endothermic).
this is important to keep track of because with these signs i can determine the reaction changes of energy and to where it's being shifted.