The standard cell potential (Eºcell) for the reaction below is +0.63 V. The cell potential

Does anyone know this

LekaFEV [45]

Answer:

Your answer would be (D

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3 0

2 years ago

Read 2 more answers

Natalia dipped two feathers in oil. Then she dipped one feather in cold water and the other feather in hot water. She swirled bo

Alika [10]

Answer:

The hot water was better for removing the oil.

Explanation:

You can see that because the mass went down more with the hot water. So, that means that more oil was taken out of the feather with hot water.

8 0

1 year ago

Which type of intermolecular attractive force operates between (part A) all molecules, (part B) polar molecules, (part C) the hy

nydimaria [60]

Answer:

PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.

PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.

PART C: Dipole dipole and in some cases hydrogen bonding operate between the hydrogen atom of a polar bond and a nearby small electronegative atom. Only if the atom bonded to it were F, O or N it would be hydrogen bonding. Otherwise it is dipole dipole.

3 0

2 years ago

A 6.000L tank at 19.2°C is filled with 18.0g of carbon monoxide gas and 10.6g of chlorine pentafluoride gas. You can assume both

Jobisdone [24]

Answer:

Total pressure: 2.89 atm

Mole fraction CO: 0.88

Partial pressure CO: 2.56 atm

Mole fraction ClF₅: 0.12

Partial pressure ClF₅: 0.33 atm

Explanation:

We should apply the Ideal Gases Law to solve this:

P . V = n . R . T

We need n, which is the total moles for the mixture

Total moles = Moles of CO + Moles of ClF₅

Moles of CO = mass of CO / molar mass CO → 18 g/28 g/mol = 0.643 mol

Moles of ClF₅ = mass of ClF₅ / molar mass ClF₅ → 10.6g/ 130.45 g/m = 0.0812 mol

0.643 mol + 0.0812 mol → 0.724 moles in the mixture

So we have the total moles so with the formula we would know the total pressure.

P . 6L = 0.724 mol . 0.082L.atm/mol.K . 292.2K

P = ( 0.724 mol . 0.082L.atm/mol.K . 292.2K) / 6L

P = 2.89 atm

Mole fraction is defined as the quotient between the moles of gas over total moles, and it is equal to partial pressure of that gas over total pressure

Moles of gas X /Total moles = Partial pressure of gas X/Total pressure

(Moles of gas X / Total moles) . Total pressure = Partial pressure of gas X

Mole fraction CO = 0.643 / 0.724 = 0.88

Partial pressure CO = 0.88 . 2.89 atm → 2.56 atm

Mole fraction ClF₅ = 0.0812 / 0.724 = 0.12

Partial pressure ClF₅ = 0.12 . 2.89 atm → 0.33 atm

5 0

2 years ago

Calculate the value deltaG°

atroni [7]

Answer:

ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.

Explanation:

4 0

2 years ago