Answer:
Iterate to improve the solution
Explanation:
apex
1st find the pOH = 14 - 4.75 = 9.25
then do 10^-9.25 = 5.62x10^-10 OH- concentration
The equation that is correct for coffee-cup calorimeter is q reaction = -q calorimeter. Details about coffee-cup calorimeter.
<h3>What is a calorimeter?</h3>
A calorimeter is an apparatus for measuring the heat generated or absorbed by either a chemical reaction, change of phase or some other physical change.
A coffee-cup calorimeter is a specific type of calorimeter that involves the absorption of heat of a reaction by water when a reaction occurs.
The enthalpy change of the reaction is equal in magnitude but opposite in sign to the heat flow for the water:
qreaction = -(qwater)
Therefore, the equation that is correct for coffee-cup calorimeter is q reaction = -q calorimeter.
Learn more about coffee-cup calorimeter at: brainly.com/question/27828855
#SPJ1
Answer:
Explanation:
Using Dalton's law of partial pressure
P total pressure = Pressure of helium + Pressure of neon + Vapor pressure of water
P = 28.3 mmHg, Pressure of helium = 381 mmHg, Vapor pressure of water at 28°C
791 mmHg - 381 mmHg - 28.3 mmHg = Pressure of neon
Pressure of neon = 381.7 mmHg
Answer:
The standard enthalpy of formation of this isomer of 
is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
The expression for the entropy change for the reaction is as follows.
![\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%3D%5B8%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28CO_%7B2%7D%29%20%2B9%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28H_%7B2%7DO%29%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28O_%7B2%7D%29%5D)
Substitute the all values in the entropy change expression.
![-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]](https://tex.z-dn.net/?f=-5104.1kJ%2Fmol%3D%5B8%28-393.5%29%2B9%28-241.8%29kJ%2Fmol%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%280%29kJ%2Fmol%5D)
Therefore, The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.
