Answer:
The equilibrium concentration of HCl is 0.01707 M.
Explanation:
Equilibrium constant of the reaction = 
Moles of ammonium chloride = 0.573 mol
Concentration of ammonium chloride = 

Initial: 0.573 0 0
At eq'm: (0.573-x) x x
We are given:
![[NH_4Cl]_{eq}=(0.573-x)](https://tex.z-dn.net/?f=%5BNH_4Cl%5D_%7Beq%7D%3D%280.573-x%29)
![[HCl]_{eq}=x](https://tex.z-dn.net/?f=%5BHCl%5D_%7Beq%7D%3Dx)
![[NH_3]_{eq}=x](https://tex.z-dn.net/?f=%5BNH_3%5D_%7Beq%7D%3Dx)
Calculating for 'x'. we get:
The expression of
for above reaction follows:
![K_c=\frac{[HCl][NH_3]}{[NH_4Cl]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHCl%5D%5BNH_3%5D%7D%7B%5BNH_4Cl%5D%7D)
Putting values in above equation, we get:



On solving this quadratic equation we get:
x = 0.01707 M
The equilibrium concentration of HCl is 0.01707 M.
Answer:
If you increase current through an electromagnet, the magnetic field around it will increase
Hope this helps!
Pretty sure it’s 13.6 , hope I get it right
Answer:
D. cathode; reduction
Explanation:
<em>Hope this helps~</em>
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Answer:
The given statement- Aldehydes and ketones are converted into alkenes by means of a direct nucleophilic addition called the Wittig reaction, is<u> True.</u>
Explanation:
The Wittig reaction converts aldehydes and ketones into alkenes through a simple nucleophilic addition. A triphenylphosphorine ylide, also known as a phosphorane, reacts with an aldehyde/ketone to produce an oxaphosphetane, a four-membered cyclic intermediate. Instead of being isolated, the oxaphosphetane decomposes spontaneously to release triphenylphosphine oxide and an alkene.
In an SN2 reaction, triphenylphosphine, a good nucleophile, reacts with a primary alkyl halide, followed by deprotonation of the carbon with a solid base, such as butyllithium, to form the ylide. In the product alkene, the carbonyl carbon and the carbon initially bound to the halogen become two carbons with a double bond.
The Wittig reaction's true worth lies in its ability to produce an alkene with a predictable structure, as the C=C bond forms exactly where the C=O bond did in the reactant, with no isomers (other than E/Z isomers) forming.
<u>Hence , the correct option is (A) True.</u>