Answer:
The mass of the precipitate that AgCl is 3.5803 g.
Explanation:
a) To calculate the molarity of solution, we use the equation:
We are given:
Mass of solute (NaCl) = 1.46 g
Molar mass of sulfuric acid = 58.5 g/mol
Volume of solution = 
Putting values in above equation, we get:
0.09982 M is the concentration of the sodium chloride solution.
b) 
Moles of NaCl = 
according to reaction 1 mol of NaCl gives 1 mol of AgCl.
Then 0.02495 moles of NaCl will give:
of AgCl
Mass of 0.02495 moles of AgCl:
The mass of the precipitate that AgCl is 3.5803 g.
Answer:
NiCO3 (s) + 2H+ (aq) → H2O (l) + CO2 (g) + Ni2+ (aq)
Explanation:
To write the complete ionic equation:
1. Start with a balanced molecular equation.
2. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
3. indicate the correct formula and charge of each ion
4. indicate the correct number of each ion
5. write (aq) after each ion
6. Bring down all compounds with (s), (l), or (g) unchanged.
Answer:
I believe Si12H26+02 is the coefficient I might be wrong, Sorry if I am.
Answer: 58.44g
Explanation: The molar mass of NaCl is 58.44g.
