Instrumental methods of analysis rely on machines.The visualization of single molecules, single biological cells, biological tissues and nanomaterials is very important and attractive approach in analytical science.
There are several different types of instrumental analysis. Some are suitable for detecting and identifying elements, while others are better suited to compounds. In general, instrumental methods of analysis are:
-Fast
-Accurate (they reliably identify elements and compounds)
-Sensitive (they can detect very small amounts of a substance in a small amount of sample)
Answer:
Based on the information, the compound is a phospholipid.
Explanation:
Phospholipids are made up of a fatty acid tail which is hydrophobic in nature and a head which comprises of phosphates that is hydrophilic in nature. Hence, phospholipids are amphiphilic compounds so they will be partially soluble in water and will allow water-soluble substances to mix with fats.
Hence, the composition of the substance described in the question confirms that is a phospholipid. As it's structure contains hydrocarbon and phosphorus and might also contain nitrogen.
Answer:
The essence including its particular subject is outlined in the following portion mostly on clarification.
Explanation:
- The energy throughout the campfire comes from either the wood's latent chemical energy until it has been burned to steam up and launch up across the campfire. The electricity generation for something like a campfire seems to be in the context including its potential chemical energy which is contained throughout the firewood used only to inflame the situation.
- The energy output seems to be in the different types of heat energy radiating across the campfire, laser light generated off by the blaze, and perhaps a little number of electrical waves, registered throughout the firewood cracking whilst they combust throughout the blaze.
and,
chemical energy ⇒ heat energy + light energy + sound energy
<u>Answer:</u> The enthalpy of the reaction for the production of 
is coming out to be -74.9 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-74.9%29%29%5D-%5B1%5Ctimes%200%29%2B%282%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-74.9kJ)
Hence, the enthalpy of the reaction for the production of is coming out to be -74.9 kJ
