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3 years ago

1. When adding strong base to a buffer, why can we assume the equilibrium value of the OH- is zero?

Chemistry

1 answer:

malfutka [58]3 years ago

6 0

Answer:

When adding strong base to a buffer, we assume the equilibrium value of the hydroxyl ion (OH-) is zero because all hydroxyl ion (OH) react with hydrogen ion (H+) forming water. So there is no hydroxyl ion (OH-) left and values of hydroxyl ion (OH-) is become zero.

Hydrogen ion is released by weak acid which is present in the buffer solution.

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When a solution forms, what interactions between particles are involved? Which are exothermic? Which are endothermic?

Blizzard [7]

Answer:

Explanation:

There are three types of interactions involved between the particles when solution are formed.

1 : Solute - solute interaction:

2 : Solute - solvent interaction:

3 : Solvent - solvent interaction:

1 : Solute - solute interaction:

It is the inter-molecular attraction between the solute particles.

2 : Solute - solvent interaction:

It involve the inter-molecular attraction between solvent and solute particles.

3 : Solvent - solvent interaction:

It involve the intermolecular attraction between solvent particles.

Solutions are formed if the intermolecular attraction between solute particles are similar to the attraction between solvent particles.

Exothermic process:

The process will exothermic when solute solvent bonds are formed with the release of energy and energy required to brake the solute-solute particles and solvent solvent particles are less.

Endothermic process:

The process will be endothermic when energy required to break the solute-solute particles and solvent solvent particles are higher than energy released when solute solvent bonds are formed .

6 0

3 years ago

A nuclide of phosphorus 3015P decays into a nuclide of silicon 3014Si. In order to satisfy charge conservation and lepton number

jeka94

In order to satisfy charge conservation and lepton number conservation the other products must be neutron.

<h3>What is conservation of mass?</h3>

The principle of conservation of mass states that, the sum of the initial mass of reactants must be equal to final mass of the products.

$_{15}^{30}P\ --- > \ _{14}^{30} Si \ + \ ^{0}_{1} n$

The balanced reaction of radioactive decay of phosphorous shows conservation of mass.

Thus, in order to satisfy charge conservation and lepton number conservation the other products must be neutron.

Learn more about radioactive decay here: brainly.com/question/1383030

#SPJ1

3 0

2 years ago

A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt

never [62]

Answer:

a) Δπ = 1.264 atm

b) W = 128 joules

c) ΔH >> W ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

R= 0.0821 Latm/kmol

T = 300 K

calling π(sw) osmotic pressure for for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W = 101.33 j/ Latm x 1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

Q₁ = heat required to bring the solution from 300 K to boiling, 373 K

ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH = 0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0

3 years ago

Write out and BALANCE the following equation: Solid Scandium metal

Rama09 [41]

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4 0

3 years ago

Indicate what happens to the concentration of Pb2+ in each half cell

Taya2010 [7]

Answer:

Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below

The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.

Explanation:

The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.

In the Left Beaker (Left half cell), their is less concentration

Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side

In the Right Beaker (right half cell), their is more concentration

Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side

Electrons will flow from Left to Right direction.

3 0

3 years ago