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vfiekz [6]
3 years ago
8

For the reaction A+B↽−−⇀C+D, assume that the standard change in free energy has a positive value. Changing the conditions of the

reaction can alter the value of the change in free energy (ΔG). Classify the conditions as to whether each would decrease the value of ΔG, increase the value of ΔG, or not change the value of ΔG for the reaction. For each change, assume that the other variables are kept constant.
A. Adding a catalystb.
B. Increasing [C] and [D]
C. Coupling with ATP hydrolysis
D. Increasing [A] and [B]
Chemistry
1 answer:
77julia77 [94]3 years ago
7 0

Explanation:

a. Adding a catalyst

no effect .( Catalyst can only change the activation energy but not the free energy).

b. increasing [C] and [D]

Increase the free energy .

c. Coupling with ATP hydrolysis

decrease the free energy value .

d.Increasing [A] and [B]

decrease the free energy.

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1.34 milligrams is the same as _____ kg and _____g.
Karolina [17]
Answer C is for kg and but it's .00134 for grams
3 0
3 years ago
Read 2 more answers
If 0.500 mol neon at 1.00 atm and 273 K expands against a constant external pressure of 0.100 atm until the gas pressure reaches
trapecia [35]

Answer:

The work done on neon = -323 J

The internal energy change= -392.84 J

The heat absorbed by neon = -69.84 J

Explanation:

Step 1: Data given

Number of moles  = 0.500 moles

Pressure  = 1 atm

Temperature  = 273 Kelvin

The pressure will change from 1.00 atm to 0.200 atm. The temperature changes from 273 to 210 Kelvin.

a) calculate the work done on neon

W = -P(V2-V1)    

⇒ with P = the pressure = 0.1 atm

⇒ with V1 = the initial volume = nRTi /Pi

⇒ with V2 = the final volume = nRTf /Pf

W = -PnR((T2/P2) -(T1/P1))

⇒ with T2 = the final temperature = 210 K

 ⇒ with T1 = the initial temperature = 273 K

 ⇒ with P2 = the final pressure = 0.200 atm

 ⇒ with P1 = the initial pressure = 1.00 atm

W = -nR (210*(0.1/0.2) - 273*(0.1/1.00))

W = -nR*(105 - 27.3)

W= -(0.500)*(8.314)*(77.7)

W = -323 J

b) calculate the internal energy change

E = (3/2)*nRT

ΔE = Ef - Ei

ΔE =(3/2)*nR(T2-T1)

⇒ with n= number of moles = 0.500 moles

⇒ with T2 =the final temperature = 210 K

⇒ with T1 = the initial temperature = 273 K

ΔE = (3/2)*(0.5)*(8.314)(210-273)

ΔE = -392.84 J

c) Calculate the heat absorbed by neon

ΔE = q + W

q = ΔE -W

⇒ with ΔE = -392.84 J

⇒ with W = -323 J

q = -392.84 J -( -323 J)

q =-392.84 J + 323 J

q = -69.84 J

4 0
3 years ago
Hydrochloric acid is the primary component of gastric juice in the stomach (see Chemistry in Action: GERD-Too Much Acid or Not E
olga nikolaevna [1]

Answer:

Acid: HCl(aq), conjugate base: Cl⁻(aq)

Base: CO₃⁻²(aq), conjugate acid: HCO₃⁻(aq)

The rewrite reaction is shown below.

Explanation:

The acid compound is the one that loses an H⁺, and the compound formed when it happens is its conjugate base. The base compound is the acceptor of H⁺, and its conjugate acid is the compound formed (Brosted-Lowry theory).

So, the acid-base pairs are:

Acid: HCl(aq), conjugate base: Cl⁻(aq)

Base: CO₃⁻²(aq), conjugate acid: HCO₃⁻(aq)

The TUMS® is an antacid, so it intends to reduce the concentration of the strong acid HCl. So, the forward reaction is favored. It can be represented with the forward arrow larger than the reversible arrow, as shown in the image below.

8 0
3 years ago
The number of atoms in a mole of any pure substance is called
GenaCL600 [577]

The number of atoms in a mole of any pure substance is called Avogadro's number.

5 0
3 years ago
What is the temperature of 0,80 mol of a gas stored in a 275 mL cylinder at 175 kPa?
love history [14]

Answer:

First, let's express pressure P in Pa and volume V in m3:

⇒P=175 kPa

⇒P=1.75×105 Pa

and

⇒V=275 ml

⇒V=2.75×10−4 m3

Then, let's solve the ideal gas law PV=nRT for temperature T:

⇒PV=nRT

⇒T=PVnR

Substituting the appropriate values into the equation:

⇒T=1.75×105×2.75×10−40.80×8.314 K

⇒T=48.1256.6512 K

∴T≈7.24 K

Therefore, the temperature is around 7.24 K.

8 0
2 years ago
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