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Answer:
At equal concentration of HBCG and BCG^-, the colour is green. This colour first appears at pH = 3.8
Explanation:
HBCG is an indicator that is prepared by dissolving the solid in ethanol.
Since
Ka=[BCG−][H3O+][HBCG]When [BCG-] = [HBCG], then Ka = [H3O+].
If pH = 3.8
Ka= [H3O+] = -antilog pH = -antilog (3.8)
Ka= 1.58 ×10^-4
Answer:
Net ionic equation for the reaction between MgCl₂ and NaOH in water:
.
Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:
.
Explanation:
Start by finding the chemical equations for each reaction:
MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.
.
MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.
.
How to rewrite a chemical equation to produce a net ionic equation?
- Rewrite all reactants and products that ionizes completely in the solution as ions.
- Eliminate ions that exist on both sides of the equation to produce a net ionic equation.
Typical classes of chemicals that ionize completely in water:
- Soluble salts,
- Strong acids, and
- Strong bases.
Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.
Take the first reaction as an example, note the coefficients:
- MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as
and
. - NaOH is a strong base. Each unit of NaOH can be written as
and
. - Mg(OH)₂ is a weak base and should not be written.
- NaCl is a salt and is soluble in water. Each unit of NaCl can be written as
and
.
.
Ions on both sides of the equation:
, and
.
Add the state symbols:
.
For the second reaction:
.
.
.
Answer:
feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6
Explanation:
Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.
Overall balance: 
Water: 
Using substitution method, we have:
= 220.77 kg/s
= 4.16 kg/s
The maximum production rate of solid crystal is
= 10*4.16 = 416 kg/s
Around evaporator:

kg/s
Around the mixing point:

Solid crystal: 
Using the last two equations, we can obtain:


kg/s
The rate of supplying fresh feed to obtain the production rate is:
= 352.5/220.77 = 1.6