Help me, I don’t understand

Which of the graphs below might represent a mixture of pure water and ice exposed to a room temperature of 3°C?

Y_Kistochka [10]

Answer:

C. Graph C

Explanation:

We have a mixture of water and ice.

At 0 °C they are at equilibrium.

water-to-ice rate = ice-to-water rate

Next, we lower the temperature to -3 °C — just slightly below freezing.

The water will slowly turn to ice.

The water-to-ice rate will be slightly faster than the ice-to-water rate.

The purple bar will be slightly higher than the blue bar.

Graph C best represents the relative rates

A. is wrong. The ice-to-water rate is faster, so the water is melting. The temperature is slightly above freezing (say, 3 °C).

B. is wrong. The two rates are equal, so the temperature is 0 °C.

D. is wrong. The water-to-ice rate (freezing) is much greater than the ice-to-water rate, so the temperature is well below freezing( say, -10 °C).

6 0

4 years ago

Calculate E o , E, and ΔG for the following cell reactions (a) Mg(s) + Sn2+(aq) ⇌ Mg2+(aq) + Sn(s) where [Mg2+] = 0.025 M and [S

Rudik [331]

E⁰(cell) = 2.24V

E(cell) = 2.246V

∆G = -433 KJ/mol

<u>Explanation:</u>

Mg(s) + Sn²⁺(aq) ⇌ Mg²⁺(aq) + Sn(s)

[Mg2+] = 0.025 M

[Sn2+] = 0.040 M

First we need the standard reduction potentials:

. . . . . . . . . . . . . . . . . E°(V)

Mg²⁺ + 2 e⁻ ⇌ Mg(s). . .−2.372

Sn²⁺ + 2 e⁻ ⇌ Sn(s) . . . −0.13

Take the more negative (or less positive in other cases) one, and write it as an oxidation:

Mg(s) ⇌ Mg²⁺ + 2 e⁻. . .+2.372 V

Combine them,

Mg(s) + Sn²⁺ ⇌ Mg²⁺ + Sn(s)

E°(cell) = +2.372 – 0.13 V = 2.24 V

To get the cell potential under the conditions given, use the Nernst Equation:

E(cell) = E°(cell) – [(0.059)/n]•logQ = 2.24 V – 0.0295 V • log [Mg²⁺]/[Sn²⁺]

Note that the solids don't appear in Q, only the concs. of the dissolved ions.

E(cell) = 2.24 V – 0.0295 V X log (0.025)/(0.040)

= 2.24 + 0.006 V ≈ 2.246 V

The concentration ratio in Q (Sn²⁺ and Mg²⁺) is too close to 1 to shift E(cell) significantly from E°(cell) given the precision I have for the Sn reduction potential.

∆G = –nFE(cell) = –2(96.485 kJ/mol•V)(2.246 V) = –433 kJ/mol

E⁰(cell) = 2.24V

E(cell) = 2.246V

∆G = -433 KJ/mol

4 0

4 years ago

(07.06 mc)

Slav-nsk [51]

Distilled water and sealed container are in equilibrium when these objects do not exchange more heat.

<h3>What is thermal equilibrium?</h3>

The expression 'thermal equilibrium' makes references to the phenomenon by which two objects interact with each other to exchange heat energy.

Objects in contact can transfer heat to reach a thermal equilibrium state where the temperature is the same for both objects.

In conclusion, distilled water and sealed container are in equilibrium when these objects do not exchange more heat.

Learn more on thermal equilibrium here:

brainly.com/question/9459470

#SPJ1

5 0

2 years ago

For the reaction 2Na + 2H20 - 2NaOH + H2 how many grams of sodium hydroxide are

ololo11 [35]

Answer:

Explanation:

Moles of sodium

=

130

⋅

g

22.99

⋅

g

⋅

m

o

l

−

1

=

5.65

⋅

m

o

l

.

Now you have the stoichiometric equation:

2

N

a

(

s

)

+

2

H

2

O

(

l

)

→

2

N

a

O

H

(

a

q

)

+

H

2

(

g

)

↑

⏐

If there are

5.65

⋅

m

o

l

of metal, clearly

5.65

⋅

m

o

l

2

dihydrogen gas are evolved......, i.e.

2.83

⋅

m

o

l

H

2

. Typically, in a question like this, you would also be asked to calculate the volume the gas occupies under standard conditions, and the

p

H

of the resultant solution (would it be high, low, neutral?).

3 0

4 years ago

Read 2 more answers

7. A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg?

USPshnik [31]

Answer:

The volume is

Explanation:

In order to solve for the volume we use the formula for Boyle's law which is

where

P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume

Since we are finding the final volume we are finding V2

Making V2 the subject we have

From the question

P1 = 300 mmHg

V1 = 300 mL

P2 = 500 mmHg

Substitute the values into the above formula and solve for the final volume obtained

That's

<h3> $V _{2} = \frac{300 \times 300}{500} \\ = \frac{90000}{500} \\ = \frac{900}{5}$ </h3>

We have the final answer as

Hope this helps you

7 0

4 years ago