Answer:
When fuels are burned in vehicle engines, high temperatures are reached. At these high temperatures, nitrogen and oxygen from the air combine to produce nitrogen monoxide. When this nitrogen monoxide is released from vehicle exhaust systems, it combines with oxygen in the air to form nitrogen dioxide.
Explanation:
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<u>Answer:</u> The 
for the reaction is 51.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical equation for the reaction of carbon and water follows:
The intermediate balanced chemical reaction are:
(1) 
( × 2)
(2) 
( × 2)
(3) 
The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B2%5Ctimes%20%5CDelta%20H_1%5D%2B%5B2%5Ctimes%20%5CDelta%20H_2%5D%2B%5B1%5Ctimes%20%28-%5CDelta%20H_3%29%5D)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-393.7%29%29%2B%282%5Ctimes%20%28-285.9%29%29%2B%281%5Ctimes%20-%28-1411%29%29%5D%3D51.8kJ)
Hence, the for the reaction is 51.8 kJ.
Fission reactions generate thermal energy
Answer:
The Barium flame is green because it is a difficult flame to excite, therefore for it to trigger a flame it is necessary that it be too excited for it to occur.
The reddish color of calcium is due to its high volatility and it is sometimes very difficult to differentiate it from strontium.the compression of these elements is due to being able to make them work during combustion
Explanation:
The flame test is a widely used qualitative analysis method to identify the presence of a certain chemical element in a sample. To carry it out you must have a gas burner. Usually a Bunsen burner, since the temperature of the flame is high enough to carry out the experience (a wick burner with an alcohol tank is not useful). The flame temperature of the Bunsen burner must first be adjusted until it is no longer yellowish and has a bluish hue to the body of the flame and a colorless envelope. Then the tip of a clean platinum or nichrome rod (an alloy of nickel and chromium), or failing that of glass, is impregnated with a small amount of the substance to be analyzed and, subsequently, the rod is introduced into the flame, trying to locate the tip in the least colored part of the flame.
The electrons in these will jump to higher levels from the lower levels and immediately (the time that an electron can be in higher levels is of the order of nanoseconds), they will emit energy in all directions in the form of electromagnetic radiation (light) of frequencies characteristics. This is what is called an atomic emission spectrum.
At a macroscopic level, it is observed that the sample, when heated in the flame, will provide a characteristic color to it. For example, if the tip of a rod is impregnated with a drop of Ca2 + solution (the previous notation indicates that it is the calcium ion, that is, the calcium atom that has lost two electrons), the color observed is brick red .
Answer:
Repeated SN2 reactions occur leading to the formation of a racemic mixture
Explanation:
S-2-iodooctane is a chiral alkyl halide with an asymmetric carbon atom. The presence of an asymmetric carbon atom implies that it can rotate plane polarized light and thus lead to optical isomerism. The two configurations of the compound are R/S according to the Cahn-Prelong-Ingold system.
However, when S-2-iodooctane is treated with sodium iodide in acetone, repeated SN2 reactions occur since the iodide ion is both a good nucleophile and a good leaving group. Hence a racemic modification is formed in the system with time hence we end up with (±)- Iodooctane.
