A gas occupies 14.3 liters at a pressure of 45.0 mm Hg. What is the volume when the pressure is increased to

When considering the relationship among standard free energy change, equilibrium constants, and standard cell potential, the equ

DedPeter [7]

Answer:

ΔG° = - RTLnK is used to find the standard cell potential given the equilibrium constant

Explanation:

for an ideal disolution:

⇒ ΔG = RT∑ni LnXi

∴ ΔG = ( μi - μi*)ni

∴ μ : chemical potential

∴ μ*: chem. potential of the pure component at T and P.

⇒ ΔG = μi - μi* = RT LnXi

for a equilibrium reaction:

⇒ ∑ νi*μi = 0

⇒ ΔGr = ΔG°+ RT Ln Kx = 0

⇒ ΔG° = - RT LnKx

4 0

4 years ago

You have 750 grams of water at 80° Celsius. Which of the following would lower the temperature of the water by 10° Celsius? (1 p

dem82 [27]

Answer:

adding 750 grams of water at 60° Celsius .

Explanation:

We can calculate the amount of heat lost from 750 grams of water at 80°C to be lowered by 10°C using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat lost by water (Q = ??? J).

m is the mass of water (m = 750.0 g).

c is the specific heat capacity of the water (c = 4.18 J/g.°C).

ΔT is the temperature difference (ΔT = final T - initial T = - 10.0°C, the temperature of water is lowered by 10.0°C).

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(- 10.0°C) = - 31350.0 J = -31.350 kJ.

Now, we can calculate the Q that is gained by the different added amounts of water:

adding 750 grams of water at 50° Celsius :

ΔT = 70.0°C - 50.0°C = 20.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(20.0°C) = 62700.0 J = 62.70 kJ.

adding 325 grams of water at 60° Celsius :

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (325.0 g)(4.18 J/g.°C)(10.0°C) = 13585.0 J = 13.585 kJ.

adding 750 grams of water at 60° Celsius :

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(10.0°C) = 31350.0 J = 31.350 kJ.

adding 1000 grams of water at 55° Celsius:

ΔT = 70.0°C - 55.0°C = 15.0°C,

∴ Q = m.c.ΔT = (1000.0 g)(4.18 J/g.°C)(15.0°C) = 62700.0 J = 62.70 kJ.

So, the right choice is:

adding 750 grams of water at 60° Celsius

8 0

4 years ago

Balance this equation. if a coefficient of "1" is required, choose "blank" for that box.

blondinia [14]

Step 1: Write Imbalance Equation

CH₃CHO + O₂   → CO₂ + H₂O

Step 2: Balance Carbon Atoms:
There are 2 carbon atoms at reactant side and one at product side. So multiply CO₂ with 2 to balance them. i.e.

CH₃CHO + O₂   → 2 CO₂ + H₂O

Step 3: Balance Hydrogen Atoms:
There are 4 hydrogen atoms at reactant side and 2 Hydrogen atoms at product side. So, multiply H₂O by 2 to balance Hydrogen on both sides. i.e.

  CH₃CHO + O₂   → 2 CO₂ + 2 H₂O

Step 4: Balance Oxygen Atoms:
There are 3 Oxygen atoms at reactant side and 6 Oxygen atoms at product side. In order to balance them multiply O₂ on reactant side by 2.5 (5/2). i.e

CH₃CHO + 5/2 O₂   →   2 CO₂ +  2 H₂O

Step 6: Eliminate Fraction:
Multiply overall equation by 2 to eliminate fraction. i.e.

2 CH₃CHO + 5 O₂   →   4 CO₂ +  4 H₂O

7 0

3 years ago

Solid lead acetate is slowly added to 75.0 mL of a 0.0492 M sodium sulfate solution. What is the concentration of lead ion requi

Firdavs [7]

Answer:

The concentration of lead ion required to just initiate precipitation is - $2.37\times10^-^5 M$