Answer:
C - raising the temperature of a gas
Explanation:
as you raise temperature, kinetic energy rises, and so does pressure
Answer:
a. Hydrocarbons have low boiling points compared to compounds of similar molar mass.
b. Hydrocarbons are hydrophobic.
d. Hydrocarbons are insoluble in water.
Explanation:
As we know that the hydrocarbons is a mix of carbon and hydrogen. In this the availability of the electronegative atom is not there that shows there is no bonding of the hydrogen plus it is dissolved. Also, the hydrocarbons is considered to be a non-polar but as compared to the water, water is a polar
In addition to this, the strong bond is no existed that shows the lower boiling points
Therefore option A, B and D are right
15% = 15 grams of solute in 100 mL solution
15 g --------------- 100 mL
?? ------------------ 1000 mL
1000 x 15 / 100 =
15000 / 100 => 150 g of solute
hope this helps!
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.
