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2 years ago

Which event occurs after erosion of earths surface

Chemistry

1 answer:

soldier1979 [14.2K]2 years ago

5 0

Answer:

Surface material breaks down into smaller pieces. Very dense particles settle faster than low-density particles. Large particles are carried away to a new location by ice and glaciers.

Explanation:

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Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine

yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

7 0

2 years ago

If all the water in 430.0 mL of a 0.45 M NaCI solution evaporates what is the mass of NaCI will remain

skad [1K]

Answer:

11.31 g.

Explanation:

Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.

M = (no. of moles of solute)/(V of the solution (L)).

∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).

∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass) = (0.45 M)(0.43 L)(58.44 g/mol) = 11.31 g.

6 0

2 years ago

Read 2 more answers

How much heat is needed to raise the temperature of 55.0 g sample of water by 65.0 oC.

sveticcg [70]

Answer: 14943.5 J

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = ?

Mass of water = 55.0g

C = 4.18 J/g°C

Φ = 65.0°C

Then, Q = MCΦ

Q = 55.0g x 4.18 J/g°C x 65.0°C

Q = 14943.5 J

Thus, 14943.5 joules of heat is needed to raise the temperature of water.

3 0

2 years ago

Read 2 more answers

Please answer with an actual answer and not just put a random word ^-^

WARRIOR [948]

Which eclipse was modeled when the large ball was between the small ball and the light?

The model is a "Lunar Eclipse" (If it was talking about the earth, then yes, it is a lunar eclipse).

Which eclipse was modeled when the small ball was between the large ball and the light?

The model is a "Solar Eclipse".

What does the large ball represent?

The earth.

What does the small ball represent?

The moon.

What does the light source represent?

The sun.

Hope this helps!~ <3

(I can't draw so sorry.)

6 0

2 years ago

In an experiment, 34.8243g of copper (II) nitrate hydrate, Cu(NO3)2•zH2O was heated to a constant mass of 27.0351g. Calculate th

Leto [7]

Answer:

1) The mass of water lost = 7.7892 grams

2) Z = 3: Cu(NO3)2*3H2O

Explanation:

Step 1: Data given

Mass of copper (II) nitrate hydrate, Cu(NO3)2•zH2O = 34.8243 grams

Mass of substance after heating = 27.0351 grams

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate mass of water

The mass of water is the mass lost after heating.

Mass water = 34.8243 - 27.0351 = 7.7892 grams of water

Step 3: Calculate moles of Cu(NO3)2

Moles Cu(NO3)2 = Mass Cu(NO3)2 / Molar mass Cu(NO3)2

Moles Cu(NO3)2 = 27.0351 grams / 187.56 g/mol

Moles Cu(NO3)2 = 0.144 moles

Step 4: Calculate moles of H2O

Moles H2O = 7.7892 grams / 18.02 g/mol

Moles H2O = 0.432 moles

Step 5: Calculate Z

z = moles H2O / moles Cu(NO3)2

Z = 0.432/0.144

Z = 3

This means we have 3 water molecules in the formula. This makes the formula ofthe hydrate: Cu(NO3)2*3H2O

5 0

2 years ago