Answer:
0.21 g
Explanation:
The equation of the reaction is;
NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)
Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles
Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles
Since the reaction is 1:1, NaCl is the limiting reactant.
1 mole of NaCl yields 1 mole of AgCl
0.00147 moles of NaCl yields 0.00147 moles of AgCl
Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol
= 0.21 g
Answer:
No.
Explanation:
No. There is 1 atom of Ca on the left and 2 Ca's on the right and 2 OH's on the left and 4 on the right.
The balanced equation is:
4OH- + 2Ca2+ ----> 2Ca(OH)2.
25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)
<h3>Explanation</h3>
The process:
.
How many moles of this process?
Relative atomic mass from a modern periodic table:
- K: 39.098;
- N: 14.007;
- O: 15.999.
Molar mass of 
:
.
Number of moles of the process = Number of moles of dissolved:
.
What's the enthalpy change of this process?
for 
. By convention, the enthalpy change 
measures the energy change for each mole of a process.
.
The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.
It would cause a drop <span>but I am not sure double check other answers </span>
Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
