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3 years ago

15

If you had a pot of boiling water that is loosing volume is that at equilibrium? What are some things you can do to a pot of boi

ling water to adjust its equilibrium?

1 answer:

ki77a [65]3 years ago

8 0

Answer:

are you rich in points brother ? or sister ?

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Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Cu2++2e−→Cu;E∘=0

Maru [420]

Answer:

Explanation:

To calculate the cell potential we use the relation:

Eº cell = Eº oxidation + Eº reduction

Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .

So the species that is going to be oxidized is the Aluminium, and therefore:

Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V

Equally valid is to write the equation as:

Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species

These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.

3 0

3 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

4 years ago

At a certain temperature the rate of this reaction is second order in NH4OH with a rate constant of 34.1 M^-1s^-1: Suppose a ves

Tom [10]

Answer:

Time = 0.929s = 0.93s (2 s.f)

Explanation:

Rate constant, k = 34.1 M^-1s^-1

Initial Concentration, [A]o = 0.100M

Time = ?

Final Concentration [A] = 0.0240M

The parameters are represented in the following equation as;

1/[A] = kt + 1/[A]o

kt = 1/[A] - 1/[A]o

kt = 1/0.0240 - 1/0.1

kt = 31.67

t = 31.67 / 34.1

t = 0.929s = 0.93s (2 s.f)

5 0

3 years ago

Which of the following does NOT involve a physical change?

jasenka [17]

Answer:mixing decomposing

Explanation:

Melting and grinding changes the physical form and mixing decomposing doesn’t

4 0

3 years ago

How many grams are 1.25 moles of potassium bromide (KBr)?

sukhopar [10]

Answer:

The SI base unit for amount of substance is the mole. one mole is equal to one mole potassium bromide or 119.0023 grams

6 0

3 years ago