Step 1:
Divide mass of each element with its M.mass in order to find out moles.
C = 63.2 g / 12 g/mol = Moles = 5.26 moles
H = 5.26 g / 1.008 g/mol = Moles = 5.21 moles
C = 41.6 g / 16 g/mol = Moles = 2.6 moles
Step 2:
Select moles of the element with least value and divide all moles of element by it,
C H O
5.26/2.6 : 5.21/2.6 : 2.6/2.6
2.02 : 2.00 : 1
Result:
Empirical Formula = C₂H₂O
Scientific questions and hypotheses come up frequently while one is engaged in investigating a scientific phenomenon such as natural geological phenomena as may occur in geological mapping in the field. For example, there may be a question does this canyon or deeply incised valley which is quite straight follow a weakness in the earth's crust like a major fault or the direction of bedding in well bedded sedimentary rocks. In a particular topographic area, some hypotheses which may be developed is that valleys follow geological structure whereas ridges follow resistant rocks like quartzites or quartz sandstones or in the ocean, points or capes may represent resistant quartz sandstones and bays may represent weak soft shales recessively weathering
Answer: Enthalpy of combustion (per mole) of
is -2657.5 kJ
Explanation:
The chemical equation for the combustion of butane follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20%5CDelta%20H%5Eo_f_%7BCO_2%28g%29%7D%29%2B%2810%5Ctimes%20%5CDelta%20H%5Eo_f_%7BH_2O%28g%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7BC_4H_%7B10%7D%28g%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7BO_2%28g%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.6%29%2B%284%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-5315kJ)
Enthalpy of combustion (per mole) of
is -2657.5 kJ
Answer:
tetrahedral geometry
<h3>CHCH2O- CH2CH3</h3>
Explanation:
There are several centers of interest. Each carbon with all single bonds is the center of a tetrahedral geometry.