Answer:
2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom
Explanation:
Given electronic configurations are :
1st: 
2nd : 
given 1st ionization energy are: 2080 kJ/mol and 496 kJ/mol
generally ionization energy of fulfilled orbital is more than half filled orbital and these two state are more stable.
therefore ionization energy of fulfilled is more than half filled orbital
hence
ionization energy of 1st atom will be very high because its orbital is fulfilled and less energy for 2nd atom so 2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom.
Answer:
Positron emission
Explanation:
Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.
A positron is usually ejected in the process together with an anti-neutrino to balance the spins.
-OH is elctron donating -C=-N is electron withdrawing -O-CO-CH3 is electron withdrawing -N(CH3)2 is electron donating -C(CH3)3 is electron donating -CO-O-CH3 is electron withdrawing -CH(CH3)2 is electron donating -NO2 is electrong withdrawing -CH2
Answer:
pH 9,8 is likely to work best for this separation
Explanation:
Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).
For lysine, PI is:
9,8
For arginine:
10,75
At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.
Thus, <em>pH 9,8 is likely to work best for this separation</em>
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I hope it helps!
Answer:
296.1 day.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).
t is the time of the reaction (t = ??? day).
a is the initial concentration of Ir-192 (a = 560.0 dpm).
(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).
<em>∴ kt = lna/(a-x)</em>
(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).
(9.365 x 10⁻³ day⁻¹)(t) = 2.773.
<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>
