Answer:
(1). CH3-CH=N-CH2-CH2-CH3.
(2). CH3-CH2-CH =N-CH2-CH3.
(3). CH3-CH2-CH2-CH=N-CH3.
(3). CH3-CH2-CH2-CH2-CH=N.
Explanation:
This is a spectroscopy question. Spectroscopy is an important part of chemistry which is used in the identification of chemical compounds.
So, without mincing words let's dive straight into the solution to the question.
Starting from values of IR given, the absorption at 1620-1680 cm^-1 shows the presence of C = N.
From the question, it is given that there is no signal showing on the spectrum at 3010 - 3090 cm^-1. The absorption at 3010 - 3090 cm^-1 shows the presence of N - H. But, it is not show on the spectrum. Hence, there is no N - H in the compound. That's, there are no Hydrogen atoms attached to Nitrogen atom.
We are given that the mass spectrum of compound A shows the molecular ion at m/z 85, therefore the molar mass of the compound is 85.
Also, from the question the elements present are carbon, Hydrogen and Nitrogen. Therefore, the compound is likely to be C5H11N.
The likely compounds are given below;
(1). CH3-CH=N-CH2-CH2-CH3.
(2). CH3-CH2-CH =N-CH2-CH3.
(3). CH3-CH2-CH2-CH=N-CH3.
(3). CH3-CH2-CH2-CH2-CH=N.
Given that a 0.130 M HCl(aq) solution costs $39.95 for 500 mL, and that KCl costs $10/ton, which analysis procedure is more cost-effective
Answer:
KCl is cost effective
Explanation:
In order to know this, we need to see how much it cost 1 g of each reactant. Let's begin with HCl
HCl:
In this case, let's calculate the moles of HCl in a 0.130 M solution and then, the mass of HCl using the molecular weight of 36.5 g/mol, to get the cost the HCl at the end using the given price:
nHCl = 0.130 moles/L * 0.5 L = 0.065 moles
mHCl = 0.065 moles * 36.5 g/mol = 2.3725 g
Cost HCl = 39.95 $ / 2.3725 g = 16.84 $/g
Conclusion, 1 g of HCl costs 16.84 $
KCl:
In this case, it's pretty obvious that 1 ton of KCl cost 10$, so, there is no need to do further calculations because 1 ton (or more than 1000 kg of the salt) it's just 10$. This is less expensive than the 16.84$ for just 1 g of HCl, so, final conclusion, KCl is more cost-effective.
Hope this helps
Draw the bridged bromonium ion that is formed as an intermediate during the bromination of this alkene. Include hydrogen atoms, nonbonding electrons, and formal charge(s) in your structure. The starting alkene is a 4 carbon chain with a double bond between carbons 2 and 3. The substituents on the alkene are on opposite sides of the alkene. This reacts with B r 2 to give the intermediate ion.
Answer:
See explanation and image attached
Explanation:
The reaction of bromine molecule with an alkene passes through a bridged intermediate known as the brominium ion.
It is a cyclic intermediate that contains a positively charged bromine ion as i have shown in the image attached.
The brominium ion is first formed during the bromininaton of alkenes.