Döbereiner grouped the known elements into <em>triads</em> (sets of three) so that
• The <em>atomic mass of the middle element</em> was approximately the average of the other two
• The <em>chemical properties of the middle element</em> were between those of the other two
• The <em>physical properties of the middle element</em> were between those of the other two
One example of a triad is Li – Na – K.
(a) Atomic mass of Na = 23.0 u
Average atomic mass of Li and K = (6.9 u + 39.1 u)/2 = 46.0 u/2 = 23.0 u
(b) Li reacts slowly with water. Na reacts rapidly. Potassium reacts violently.
(c) Melting point of Na = 371 °C.
Average melting point of Li and K = (454 °C + 330 °C)/2 = 784 °C/2
= 392 °C
The answer is 35.4335
Hope this helped! (Plz mark me brainliest!)
an element's name, chemical symbol, atomic number, atomic mass.
IDK what you are even asking for
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
