The product that you obtain if you evaporate the water from the NaoH layer prior to acidifying the layer is salt of <u> para - tery-butyphenol</u>
<u><em>Explanation</em></u>
- Para-tery-butyphenol can be isolate after extracting it into NaOH solution through carrying out
- heating and then cooling,
- then 3M of HCl is added to acidify.
- Then ice bath is done if needed
- finally the precipitate is filtered.
- Para tery butyphenol can be used for production of alkylphenolic.
Ionization enthalpy, IE, is also called ionization potential is the ability to remove the electron from the neutral gaseous atom. There is a trend observed in the periodic table for the IE value. As we go from left to right in a period, IE vale increases. While moving from top to bottom in a group, IE value decreases.
- The phenomenon of unexpected drop in IE1 values between Groups 2 and 13, in period 2 and period 4 is due to the introduction of d-orbitals in the case of period 4 elements.
- While moving in the period, there is the constant addition of electrons in the nucleus. The shell sie remains constant while electron pull increases from the nucleus, this leads to a reduction in the size of the atom. As the size decreases, it is difficult to remove the electron from the atom, and thus IE value increases in the case of period 2.
- When we study the case of period 4, there is an introduction of d-electrons. As the inner shell electron increases, there is an increase in the shielding effect. This shielding effect tends to decrease the nuclear attraction between the nucleus and outermost electrons. Ultimately this decreases the IE value in the fourth period. Such a phenomenon is absent in the case of group 2 elements.
- If we speak in terms of orbital energy, the IE value decreases while moving from top to bottom in the period. This is due to the fact that, as we go down in the periodic table, the number of shells increases, and the outermost electron is too far from the nuclear attraction, therefore it can be ejected out easily. This marks a decrease in IE value.
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Answer:
1.20 M
Explanation:
Convert grams of Na₂CO₃ to moles. (50.84 g)/(105.99 g/mol) = 0.4797 mol
Molarity is (moles of solute)/(liters of solvent) = (0.4797 mol)/(0.400 L) = 1.20 M
Answer:
aplha particle is the answer