Answer:
6.096799125kg
Explanation:
According to the question, three different samples weighed using different types of balance had masses: 0.6160959 kg, 3.225 mg, and 5480.7 g.
Based on observation, the mass units in the three measurements are different but must be uniform in order to find the total mass. Hence, we need to convert to the standard unit (S.I unit of mass), which is kilograms (kg)
Since 1kg equals 1,000,000mg
Hence, 3.225mg will be 3.225/1000000
= 0.000003225kg
Also, 1kg equals 1000g
Hence, 5480.7g will be 5480.7/1000
= 5.4087kg
Hence, the total mass of the three samples (now in the same unit) are:
5.4807kg + 0.000003225kg + 0.6160959 kg
= 6.096799125kg
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
<h2>Please mark me as brainliest please yaa!!!^_^</h2>
Answer:
Element Lithium
Explanation:
The element with the highest second ionization energy is lithium. It belongs to the alkaline metal group I.e group one metals
It has the highest second ionization energy because it is very difficult to remove the electron from the 1s orbital.
Its atomic number is 3. The electronic configuration is 1s2 2S1