Answer:
d and e - Sodium and antimony
Explanation:
The atomic numbers remain the same, while the mass numbers change (because neutrons are being added or taken away).
sodium has an atomic number of 19 and a mass number of 39 - in d, it has an atomic number of 19 but a mass number of 40. therefore, it is an isotope
antimony has an atomic number of 51 and a mass number of 121.60 - in e, it has an atomic number of 51, but a mass number of 123. therefore, it is an isotope
This uses something called <span>Le Chatelier's principle. It states essentially that any stress put upon a system will be corrected.
In more simple terms, it means that in an equilibrium, such as the equation N2(g) + 3H2(g) <=> 2NH3(g), removing a reactant will cause the system to create more of said reactant to compensate for its loss, or adding excess reactant will cause the system to remove some of the added reactant. For future reference, the same principle applies to products in an equilibrium as well.
In this case, hydrogen gas is a reactant, and hydrogen is being removed. According to </span><span>Le Chatelier's principle, the system will shift to create more hydrogen gas. In essence, it will shift in the direction of the hydrogen gas, so there will be a shift toward the reactants.
To clear something up, Keq will not change, as it is a constant value with constant conditions (such as temperature, pressure, etc.).</span>
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
Answer:
a.reducing agent reduces other elements and make itself oxidized
b.Oxidizing agent reduces itself and make other element oxidized
c.Oxidation state is apparent charge on an atom
Explanation:
Answer:
100g / (5.2g/cm3)
= 100g / (5.2g / 1cm3)
= 100g x 1cm3 / 5.2 g
= 19.2 cm3
Since 1 cm3 = 1 ml, your answer is 19.2 ml.
