Answer:
Yes. Ca²⁺(aq) + SO₄²⁻(aq) ⟶ CaSO₄(s)
Step-by-step explanation:
The possible reaction is
CaI₂ + Na₂SO₄ ⟶ CaSO₄ + 2NaI
To predict a reaction, you must know <em>the solubility rules</em>.
The important ones for this question are
- Salts containing Na⁺ are <em>soluble</em>. Thus, NaI is soluble.
- Most sulfates are soluble, but CaSO₄ is not. Thus, CaSO₄ is a precipitate, and a reaction occurs.
Molecular equation
CaI₂(aq) + Na₂SO₄(aq) ⟶ CaSO₄(s) + 2NaI(aq)
Ionic equation
Ca²⁺(aq) + <u>2I⁻(aq)</u> + 2Na²⁺(aq) + SO₄²⁻(aq) ⟶ CaSO₄(s) + 2Na⁺(aq) + 2I⁻(aq)
Net ionic equation
Cancel all ions that appear on both sides of the reaction arrow (underlined).
Ca²⁺(aq) + <u>2I⁻(aq) </u>+ <u>2Na²⁺(aq)</u> + SO₄²⁻(aq) ⟶ CaSO₄(s) + <u>2Na⁺(aq)</u> + <u>2I⁻(aq)
</u>
The net ionic equation is
Ca²⁺(aq) + SO₄²⁻(aq) ⟶ CaSO₄(s)
Answer: The equilibrium constant is 
Explanation:
Initial concentration of
= 0.095 M
The given balanced equilibrium reaction is,

Initial conc. 0.095 M 0 M
At eqm. conc. (0.095-x) M (2x) M
Given : 2x = 0.0055
x = 0.00275
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :


Thus the equilibrium constant is 
<em>ANSWER:</em>
Objects that have the same charge, both positive or both negative, repel each other, and those with opposite charges attract each other. ... Similar to electric charges, opposite poles attract, and like poles repel. The stronger the magnets and the closer together they are, the stronger the magnetic force between them.
1 liter = 2.11 pints
8 pints = 1 gallon
4900 L * (2.11 pints/1 L) * (1 gallon/8 pints) = 1292.375 gallons
4900 has 2 significant figures, so our final answer should have two. Your final answer is 1300 gallons.
Answer:
a. Minimum 1.70 V
b. There is no maximum.
Explanation:
We can solve this question by remembering that the cell potential is given by the formula
ε⁰ cell = ε⁰ reduction - ε⁰ oxidation
Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the oxidized species 0.80 V, thus
ε⁰ reduction - ε⁰ oxidation ≥ ε⁰ cell
Since ε⁰ oxidation is by definition the negative of ε⁰ reduction , we have
ε⁰ reduction - ( 0.80 V ) ≥ 0.90 V
⇒ ε⁰ reduction ≥ 1.70 V
Therefore,
(a) The minimum standard reduction potential is 1.70 V
(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V