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Nikitich [7]
3 years ago
13

Which of the followings is true about G0'? A. G0' can be determined using Keq' B. G0' indicates if a reaction can occur under no

n-standard conditions C. G0' is always positive D. G0' is determined at pH 7 and temperature 298 K
Chemistry
1 answer:
nekit [7.7K]3 years ago
5 0

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

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The volume of a sphere is given by the equation =4/33, where is the radius. Calculate the volume of a sphere with a radius of 34
sineoko [7]

Answer:

its 45 because of 49 og the

5 0
3 years ago
C3H8 and CH4 are 2 compounds made from the same elements. This is an example of which law? A) Law of Conservation of Mass b) Law
Sphinxa [80]

C3H8 and CH4 are two compounds made from the same two elements, C and H. The ratios of C and H for both are round numbers.

Law of multiple proportions states that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

So the ans is C) Law of Multiple Proportions

6 0
3 years ago
The amount of 217 mg of an isotope is given by A(t) = 217 € -0.0171, where t is time in years since the initial amount of 217 mg
Maru [420]

The amount left after 20 years = 154.15 mg

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually radioactive elements have an unstable atomic nucleus.  

The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles  

The decay formula for isotope :

\tt \large{\boxed{\bold{A(t)=217e^{-0.0171t}}}

Then for t=20 years, the amount left :

\tt A(t)=217e^{-0.0171\times 20}\\\\A(t)=154.15~mg

4 0
3 years ago
Read 2 more answers
If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?
IceJOKER [234]

Answer:

n_{H_2O}=1.5molH_2O

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O

Best regards.

5 0
3 years ago
Students in a chemistry class are drawing molecular models based on molecular formulas they are given by the teacher.
Karo-lina-s [1.5K]
Sorry I need this marks
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3 years ago
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