Answer:
(chemical) is that what you where asking
Explanation:
can i have a brainlist
Explanation:
As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
..........(1)
..............(2)
The final reaction is as follows:
.............(3)
Therefore, adding (1) and (2) we get the final equation (3) and value of at 298 K will be as follows.
= +
= -314 kJ + (-80) kJ
= -394 kJ
Thus, we can conclude that at 298 K for the given process is -394 kJ.
Answer:
C) SN2 and E2
Explanation:
For this question, we have analyzed the <u>substrate</u> and the <u>base/nucleophile</u>. The substrate, in this case, is 1-iodohexane and the base/nucleophile is potassium tert-butoxide.
<u>Substrate</u>
<u />
In the 1-iodohexane the iodide "I" is bonded to a primary carbon (carbon 1). Therefore we will have a <u>primary substrate</u>. If we have a primary substrate an Sn1 can not take place. We can not have a <u>primary carbocation</u> due to this instability. So, we can disccard options A) and B).
<u>Base/nucleophile</u>
<u />
In the potassium tert-butoxide we have an ionic compound. A positive charge is placed in the potassium atom a negative charge is placed in the oxygen of the ter-butoxide ion. So, we will have a <u>strong base</u> (a molecule with the ability to remove electrons) and a <u>strong nucleophile</u> (a molecule with ability to bond with an electrophile). With all this in mind, w<u>e can not have an E1 reaction</u>.
With both analyses, the answer is C).
See figure 1
I hope it helps!
Answer:
a) K = [ CO2(g) ]
⇒ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
Explanation:
a) CaCO3(s) ↔ CaO(s) + CO2(g)
⇒ K = [ CO2(g) ]
∴ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) H2(g) + F2(g) ↔ 2 HF(g)
⇒ K = [ HF(g) ] ² / [ F2(g) ] * [ H2(g) ]
⇒ Kp = PHF² / PF2 * PH2
for ideal gas:
PV = RTn
⇒ P = n/V RT = [ ] RT
⇒ Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same.
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
Answer:
<em>Butane has only two isomers and pentane has just three, but some hydrocarbons have many more isomers than these. As you increase the number of carbon atoms in a hydrocarbon, the number of isomers quickly increases.</em>
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