Answer: The workdone W = 505J
Explanation:
Applying the pressure-volume relationship
W= - PΔV
Where negative sign indicates the power is being delivered to the surrounding
W = - 1.0atm * ( 5.88 - 0.9)L
= - 1.0atm * (4.98)
W = -4.98 atmL
Converting to Joules
1atmL = 101.325J
-4.98atmL = x joules.
Work done in J = -4.98 * 101.325
W= -505J
Therefore the workdone is -505J
Answer:
The K sp Value is 
Explanation:
From the question we are told that
The of
is = 122.5 g/ mol
The mass of
dissolved is 
The volume of solution is 
The number of moles of
is mathematically evaluated as

Substituting values


Generally concentration is mathematically represented as
For


The dissociation reaction of
is

The solubility product constant is mathematically represented as

Since there is no ionic reactant we have
![K_{sp} = [k^+] [ClO_3^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5Bk%5E%2B%5D%20%5BClO_3%5E-%5D)



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:
.
.
.
The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:
ΔS = nCln(T₂/T₁)
n = number of moles
C = molar heat capacity
T₂ = final temperature
T₁ = initial temperature
We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:
[(3 moles)(273.15 K) + (1 mole)(373.15 K)]/(4 moles) = 298.15 K
Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:
ΔS = (3 moles)(75.3 J/Kmol)ln(298.15/273.15)
ΔS = 19.8 J/K
ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
ΔS = -16.9 J/K
Now we combine the entropy change of each portion of water to get the total entropy change for the system:
ΔS = 19.8 J/K + (-16.9 J/K)
ΔS = 2.9 J/K
The entropy change for combining the two temperatures of water is 2.9 J/K.