Answer:
the molecular weight of C12H22O11 is calculated as
12x12 + 22x1 + 11x 16
= 144 + 22 + 176
=342g/mol
the moles is 0.50 as given ...
to get grams now
342 x 0.5
= 171g
Explanation:
Answer:
Explanation:
From the combustion of carbon, the reactions occurring in limited oxygen conditions are:
If it occurs in excess, then any leftover CO changes to CO2. i.e.
---- (1)
----- (2)
From (1), the enthalpy change is:
![\Delta H_{rxn1} = \Delta H^0_{fCO_2(g)} - ( \Delta H^0_{f C(graphite)}+ \Delta H^0_{fCO_2(g)}](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn1%7D%20%3D%20%5CDelta%20H%5E0_%7BfCO_2%28g%29%7D%20-%20%28%20%5CDelta%20H%5E0_%7Bf%20C%28graphite%29%7D%2B%20%5CDelta%20H%5E0_%7BfCO_2%28g%29%7D)
![\Delta H_{rxn1} =-393.5 \ kJ/mol -(0+0)](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn1%7D%20%3D-393.5%20%5C%20kJ%2Fmol%20-%280%2B0%29)
![\Delta H_{rxn1} =-393.5 \ kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn1%7D%20%3D-393.5%20%5C%20kJ%2Fmol)
From (2), the enthalpy change is:
![\Delta_{rxn2} = \Delta H^0_{fCO_2(g)} - ( \Delta H^0_{fCO(g)} + \dfrac{1}{2} \Delta H^0_{fO_2(g)})](https://tex.z-dn.net/?f=%5CDelta_%7Brxn2%7D%20%3D%20%5CDelta%20H%5E0_%7BfCO_2%28g%29%7D%20-%20%28%20%5CDelta%20H%5E0_%7BfCO%28g%29%7D%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%20%5CDelta%20H%5E0_%7BfO_2%28g%29%7D%29)
![\Delta_{rxn2} = -393.5 \ kJ/mol -(-110.5 + \dfrac{1}{2}(0))](https://tex.z-dn.net/?f=%5CDelta_%7Brxn2%7D%20%3D%20-393.5%20%5C%20kJ%2Fmol%20-%28-110.5%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%280%29%29)
![\Delta_{rxn2} = -283.0 \ kJ/mol](https://tex.z-dn.net/?f=%5CDelta_%7Brxn2%7D%20%3D%20-283.0%20%5C%20kJ%2Fmol)
Subtracting (2) from (1), we get:
![C(graphite) + O_{2(g)} \to CO_{2(g)} \ \ \ \Delta H_{rxn} = -393.5 \ kJ/mol}](https://tex.z-dn.net/?f=C%28graphite%29%20%2B%20O_%7B2%28g%29%7D%20%5Cto%20CO_%7B2%28g%29%7D%20%5C%20%5C%20%5C%20%20%5CDelta%20H_%7Brxn%7D%20%3D%20-393.5%20%5C%20kJ%2Fmol%7D)
![CO_{(g)} + \dfrac{1}{2} O_2(g) \to CO_{2(g)}} \ \ \ \Delta H _{rxn2} = -283.0 \ kJ/mol](https://tex.z-dn.net/?f=CO_%7B%28g%29%7D%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%20O_2%28g%29%20%5Cto%20CO_%7B2%28g%29%7D%7D%20%5C%20%5C%20%5C%20%20%5CDelta%20H%20_%7Brxn2%7D%20%3D%20-283.0%20%5C%20kJ%2Fmol)
<u> </u>
![C(graphite) + O_{2(g)} \to CO (g) + \dfrac{1}{2}O_{2(g)} \ \ \ \Delta H_{rxn} = -110.5 \ kJ/mol](https://tex.z-dn.net/?f=C%28graphite%29%20%2B%20O_%7B2%28g%29%7D%20%5Cto%20CO%20%28g%29%20%2B%20%5Cdfrac%7B1%7D%7B2%7DO_%7B2%28g%29%7D%20%20%5C%20%20%20%5C%20%5C%20%20%20%5CDelta%20H_%7Brxn%7D%20%3D%20-110.5%20%5C%20kJ%2Fmol)
![C(graphite) + \dfrac{1}{2} O_{2(g)} \to CO (g) \ \ \ \Delta H_{rxn} = -110.5 \ kJ/mol](https://tex.z-dn.net/?f=C%28graphite%29%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%20O_%7B2%28g%29%7D%20%5Cto%20CO%20%28g%29%20%20%5C%20%20%20%5C%20%5C%20%20%20%5CDelta%20H_%7Brxn%7D%20%3D%20-110.5%20%5C%20kJ%2Fmol)
The enthalpy change ΔH of the reaction = -110.5 kJ/mol
The volume occupied by 2.00 moles of nitrogen gas at the same temperature and pressure will be
0.500 moles = 11.2 Liters
what about 2 moles =? liters
by cross multiplication
= 11.2 liters x 2moles/ 0.500 moles = 44.8 liters
The iterative process is the practice of building, refining, and improving a project, product, or initiative. Teams that use the iterative development process create, test, and revise until they're satisfied with the end result.
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I think it’s going to be 8, which is my best guess. Good luck