Answer:
The mole fraction of ethanol is 0.6. A 10 mL volumetric pipette must be used for to measure the 10 mL of ethanol. The vessel should be clean and purged.
Explanation:
For calculating mole fraction of ethanol, the amount of moles ethanol must be calculated. Using ethanol density (0.778 g/mL), 10 mL of ethanol equals to 7.89 g of ethanol and in turn 0.17 moles of ethanol. The same way for calculate the amount of water moles (ethanol density=0.997 g/mL). 2 mL of water correspond to 0.11. The total moles are: 0.17+0.11=0.28. Mole fraction alcohol is: 0.17/0.28=0.6
OK in the case of hydrazine 14 grams of nitrogen combine with 2 gram of hydrogen and with ammonia 14 grams combine with 3 grams of hydrogen.
Ratio 2:3
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.
The energy at n level of hydrogen atom energy level =13.6/n^2
substiture the respective n values in the equation above and find the difference in the energy levels
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