<h3>
Answer:</h3>
Li₂S(aq)+Pb(NO₃)₂(aq)
→ PbS(s) + LiNO₃(aq)
K2S(aq)+Sn(NO₃)₄(aq→ SnS(s) + KNO₃(aq)
<h3>
Explanation:</h3>
- According to solubility rules, metal sulfides are insoluble except Calcium sulfide(CaS), Magnesium sulfide(MgS), Barium sulfide(BaS), and those of potassium, sodium, and ammonium.
- Therefore, when all the other sulfides are formed during a reaction they form a precipitate that is shown to be in solid-state.
- Precipitates are formed during precipitation reactions when cations and anions combine to form a compound that is insoluble in water.
In our case; In the equations given, some reactions will take place to form a precipitate while others will not occur.
That is;
- Na₂S(aq)+KCl(aq) → No reaction
- Li₂S(aq)+Pb(NO₃)₂(aq) → PbS(s) + LiNO₃(aq)
- Pb(ClO₃)₂(aq)+NaNO₃(aq) → No reaction
- AgNO₃(aq)+KCl(aq) →AgCl(s) + KNO₃(aq)
- K2S(aq)+Sn(NO₃)₄(aq→ SnS(s) + KNO₃(aq)
The first reaction will not occur.
The second equation is a precipitation reaction that forms lead(ii)sulfide which is a black precipitate.
The third reaction will not take place.
The fourth reaction will be a precipitation reaction forming silver chloride precipitate.
The fifth equation is also a precipitation reaction that forms tin(ii) sulfide, SnS which is a black precipitate.
Therefore; the equations that answers our question are;
- Li₂S(aq)+Pb(NO₃)₂(aq) → PbS(s) + LiNO₃(aq)
- K2S(aq)+Sn(NO₃)₄(aq→ SnS(s) + KNO₃(aq)
Answer:
Any matter considered to be a fuel contains chemical energy
Energy has to be provided to initiate the reaction.
So yes. It is
Answer:
0.02498 L
Explanation:
CaCO₃ + 2 HCl ➔ CaCl₂ + H₂O + CO₂
You have to use stoichiometry. According to the chemical equation, for every one mole of CaCO₃, two moles of HCl is needed for the reaction to occur. Before you can use this relation you need to convert grams of CaCO₃ to moles. To convert, you need to use the molar mass.
Molar mass of CaCO₃ = 100.086 g/mol
(0.2500 g)/(100.086 g/mol) = 0.002498 mol CaCO₃
Now using the relation of 1 mol of CaCO₃ for every 2 mol of HCl, convert moles of CaCO₃ to moles of HCl.
(0.002498 mol CaCO₃) (2 mol HCl)/(1 mol CaCO₃) = 0.004996 mol HCl
Since the molarity of the solution of HCl is 0.200 M (mol/L), you have to divide the amount of moles needed by the molarity of the solution.
(0.004996 mol HCl)/(0.200 M) = 0.02498 L
You will need 0.02498 L to react with 0.2500 g of CaCO₃.
First fine the formula weight for <span>CaCO₃ by adding the atomic mass of each atom - Ca (40.078g) + C (12.0107g) + O * 3 (15.9994g * 3) = 100.0869g
Take the mass of each individual atom type, divide by the overall mass (100.0869g) and multiply by 100 to find the percent composition of each atom type.
(40.078g)/(100.0869g) * 100 = 40.042%
(12.0107g)/(100.0869g) * 100 = 12.0003%
(15.9994g * 3)/(100.0869g) * 100 = 47.9565%</span>