Partial pressure is the amount of pressure or force that is exerted by the atoms into the outer environment. it is dependent on the temperature and pressure of the present surroundings. in this case, we are asked in this problem to determine the partial pressure of oxygen at 16oC and 1 atm. We have to look into a solubility data table commonly found in handbooks and determined via experiments and correlations. According to literature, the value of the partial pressure is equal to 0.617 mM.This is under the assumption that the salinity of the water in which oxygen is dissolved is equal to zero.
The whole Activity , poem and paragraph is missing in the question.
Answer:
(1) Liquid A
(2) Solid A
Explanation:
Using this part of the given poem
Substances and mixtures behave differently,
During boiling and melting most especially
Boiling point of substance is fixed while mixture is not
Substance melts completely but mixture does not
The boiling point of the Pure substance remain fixed after reaching its boiling point this is shown by Liquid A
Solid A is melting completely so Solid A is a pure substance.
<span>The mass (in grams) of 8.45 x 10^23 molecules of dextrose is 252.798g
Working:
Mw. dextrose is 180.16 g/mol
therefore 180.16 grams dextrose = 1 mole
therefore 180.16 grams dextrose= 6.022x10^23 molecules (Avogadro's number)
We have 8.45 x 10^23 molecules of dextrose.
Therefore, (180.16 divided by 6.022x10^23) times 8.45x10^23
gives the mass (in grams) of 8.45 x 10^23 molecules of dextrose;
252.798.</span>
Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]](https://tex.z-dn.net/?f=-1.0940%5Ctimes%2010%5E4%3D%5B%2816%5Ctimes%20-393.5%29%2B%2818%5Ctimes%20-285.8%29%5D-%5B%2825%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelat%20H_f%7BC_8H_%7B18%7D%28l%29%7D%5D)
Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
