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zavuch27 [327]
3 years ago
13

g If attempting to dissolve both silver bromide and silver chloride in aqueous solution through complex ion formation, which dat

a will be the most relevant?
Chemistry
1 answer:
ziro4ka [17]3 years ago
7 0

Answer:

Kf

Explanation:

The stability constant Kf of a given complex specie is an equilibrium constant that represents the formation of that particular complex specie in solution. It measures the strength of the interaction between the ligands and metal that form the particular complex specie. The magnitude of Kf shows how easily a complex specie is formed in solution.

Hence if I want to dissolve the bromides or chlorides of silver which are ordinarily insoluble in water by means of complex formation, the magnitude of the stability constant for each particular complex specie is important as it gives information regarding the thermodynamic feasibility of the process.

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What causes ionic bonding between two atoms?
Marrrta [24]

Answer:

Ionic bonding happens when an atom of an element gives one or more of its electrons to the other element's atom..it usually takes place between metal and non metal atoms...like in NaCl, Na gives its valence electron to chlorine and completes its own octet. Chlorine accepts the electron and completes its own octet too...but now both the atoms have an opposing charge and hence they attract each other to form an IONIC bond.

Ionic bonds are the strongest of the bonds...here complete transfer of electrons takes place unlike covalent bonds.

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1 year ago
A filament for a light bulb needs to conduct electricity. Which of the elements listed below might be useful as a light bulb fil
horrorfan [7]

Answer:

A. tungsten

Explanation:

Tungsten is a material which high melting point ie. does not melt easily incase of high temperature

8 0
3 years ago
Read 2 more answers
Silver bromide (AgBr) has a solubility of 7.07 × 10^-7 mol/L
kirill115 [55]

a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)

b.  Ksp AgBr = s²

c. 5 x 10⁻¹³  mol/L

<h3>Further explanation</h3>

Given

solubility AgBr = 7.07 x 10⁻⁷ mol/L

Required

The dissolution reaction

Ksp

The solubility product constant

Solution

a. dissolution reaction of AgBr

AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)

b. Ksp

Ksp AgBr  = [Ag⁺]  [Br⁻]

Ksp AgBr = (s) (s)

Ksp AgBr = s²

c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³  mol/L

7 0
2 years ago
What happens when energy is added to each state?
e-lub [12.9K]

Answer:

One change of state happens when you add energy to the substance. This change of state is called melting. By adding energy to the molecules in a solid the molecules begin to move quicker and can break away from the other molecules. ... The temperature at which a substance goes from a solid to a liquid is it melting point.

5 0
3 years ago
Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemogl
Alex73 [517]

Without wasting much of our time, Here is the correct question.

Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) -------> hemoglobin O2(aq) is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol⁻¹ s⁻¹. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10⁻⁹ M and that of oxygen is 5 × 10⁻⁵M.

Answer:

4 × 10⁻⁶ M s⁻¹

Explanation:

The equation for the reaction between Hemoglobin molecules in blood that binds with oxygen molecule can be represent by:

hemoglobin_{(aq)  +  O_{2(aq)   ---------> hemoglobin.O_{2(aq)

Now, we are also being told to calculate only!, the  initial rate at which oxygen will be bound to hemoglobin.

So, If it is first order in hemoglobin and also first order in Oxygen molecule at the initial rate of the the reaction, therefore, the rate  for the reaction can be expressed as :

rate = k [hemoglobin_{(aq)}][O_{2(aq)}]

Let's not forget that we are so given some parameters;

where

k (rate constant) = 4 × 10⁷ L mol⁻¹ s⁻¹

[ hemoglobin_{(aq) ] = 2 × 10⁻⁹ M

[  O_{2(aq)  ]  =  5 × 10⁻⁵ M

Substituting our data given into the above rate formula, we have:

rate = (4 × 10⁷ L mol⁻¹ s⁻¹) × (2 × 10⁻⁹ M) × (5 × 10⁻⁵ M)

rate = 4 × 10⁻⁶ M s⁻¹     ( given that 1 M = 1 mol L⁻¹ )

∴ the initial rate at which oxygen will be bound to hemoglobin = 4 × 10⁻⁶ M s⁻¹

7 0
3 years ago
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