Answer:
- <em>The net ionic equation is: </em><u>Ag⁺ (aq) + Cl ⁻ (aq) → AgCl (s)</u>
Explanation:
<u>1) Start by writing the total ionic equation:</u>
The total ionic equation shows each aqueous substance in its ionized form, while the solid or liquid substances are shown with their chemical formula.
These are the ionic species:
- AgF (aq) → Ag⁺ (aq) + F⁻ (aq)
- NH₄Cl (aq) → NH₄⁺ (aq) + Cl ⁻ (aq)
- NH₄F(aq) → NH₄⁺ (aq) + F⁻ (aq)
Then, replace each chemical formula in the chemical equation by those ionic forms:
- Ag⁺ (aq) + F⁻ (aq) + NH₄⁺ (aq) + Cl ⁻ (aq) → AgCl (s) + NH₄⁺ (aq) + F⁻ (aq)
That is the total ionic equation.
<u>2) Spectator ions:</u>
The ions that appear in both the reactant side and the product side are considered spectator ions (they do not change), and so they are canceled.
In our total ionic equation they are F⁻ (aq) and NH₄⁺ (aq).
After canceling them, you get the net ionic equation:
<u>3) Net ionic equation:</u>
- Ag⁺ (aq) + Cl ⁻ (aq) → AgCl (s) ← answer
For the equation that we are given:
→ 
We see that on the reactants side we have:
Ca: 1 mol
O: 2 mol
H: 2 mol
And then we see that on the products side we have:
Ca: 1 mol
O: 2 mol
H: 2 mol
Therefore since we have equal moles of each of the elements on both sides of the equation, this equation is balanced how it is.
So the correct coefficients for this equation are: B) 1, 1, 1.
Answer:
The mass was there all along, it was just in the air. The weight of the oxygen from the air is not weighed in the beginning, only at the end as part of the product, making it seem like there is a total mass change.
Answer:
The answer to your question is the letter B) 1.53 M
Explanation:
Data
Molarity = ?
volume = 225 ml
mass of KNO₃ = 34.8 g
molar mass = 101.1 g/mol
Process
1.- Convert the ml to liters
1000 ml ------------------ 1 l
225 ml ------------------- x
x = (225 x 1) / 1000
x = 0.225 l
2.- Convert the mass of KNO₃ to moles
101.1 g ---------------------- 1 mol
34.8 g --------------------- x
x = (34.8 x 1) / 101.1
x = 0.344 moles
3.- Calculate the Molarity
Molarity = moles/volume
-Substitution
Molarity = 0.344/ 0.225
-Result
Molarity = 1.53 M