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:
4 moles
Explanation:
From the equation 1 mole of C6H1206 produces 6 moles of CO2.
Therefore the answer is 24/6 = 4 moles of C6H1206.
Answer:
should be A)Melts.Because magma is liquid i think
Fe2+:ferrous ion; fe3+ ferric ion are the one which are correctly named(Answer A) Sn2+ are called stannous ion while Sn4+ are known as stannic ions. Pb2+ are called lead (II) ions while Pb4+ are known as lead (IV) ion Co2+ are known was cobalt (II) ions while Co3+ are known as cobalt(IIi) ion.
<u><em>The process of how we would obtain </em></u><u><em>ethanal</em></u><u><em> </em></u><u><em>free</em></u><u><em> from ethanol is described in the explanations below. </em></u>
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- In Chemistry, Ethanol undergoes oxidation in the presence of sodium dichromate plus sulphuric acid to yield ethanal and water.
The procedure for achieving this in the laboratory is as follows;
- Step 1; Measure a quantity of a solution of sodium dichromate acidified in a dilute sulphuric acid and pour into a test tube.
- Step 2; Add excess <em>ethanol</em>. This is because if we don't do so there will be plenty of oxidizing agent to carry out a second operation which changes the aldehyde to ethanoic acid. However, we need only the aldehyde.
- Step 3; When the aldehyde ethanal begins to form which will be evident by the change in the colour of solution from <em>orange to green</em>, then the mixture should be distilled from the test tube and tbethe aldehyde collevted so that it doesn't undergo additional oxidation into ethanoic acid.
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