The shot answer for this question is (a) True
Answer:
Strong broad peak around 3200-3600 cm-1 should be present
Strong peak around around 1700 cm-1 should be absent
Explanation:
Infrared spectroscopy is an analytical technique which is used for molecular structure characterization by identifying the functional groups present in a given molecule based on the absorption wavelength (or wavenumber).
In an IR spectrum the carbonyl group is associated with the C=O stretch which occurs as a strong peak around around 1700 cm-1. For alcohol the -corresponding O-H stretching frequency occurs as a strong broad peak between 3200-3600 cm-1.
Therefore, in the case of estradiol the presence a strong broad peak in the 3200-3600 cm-1 and the absence of the peak at around 1700 cm-1. would suggest that the transformation is complete.
<span>Errors can come from various sources: the obtainer of the data, the measuring instrument, the setting and etc. Errors are what makes your measurement invalid and unreliable. There are two types of error which is called the systematic error and the random error. Each error has different sources. Words that were mentioned –invalid and unreliable are very important key aspects to determine that your measure is truly accurate and consistent. Some would recommend using the mean method, doing three trials in measuring and getting their mean, in response to this problem.</span>
Answer:
Explanation:
If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.
Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.
However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
The attached diagrams portraying this notions is shown in the attached file below.
Answer;
K+ and NO3- ions
Explanation;
The main ions remaining are K+ and NO3- ions after pbi2 precipitation is complete.
However; There will always be tiny amounts of Pb2+ and I- ions, but most of them are in the solid precipitate.
