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.
Yes chloride is an element.... hope this helps!!
Answer:
ΔG = 18KJ/mol
Explanation:
Given data:
ΔS = 0.09 Kj/mol.K
ΔH = 27 KJ/mol
Temperature = 100 K
ΔG = ?
Solution:
Formula:
ΔG = ΔH - TΔS
ΔH = enthalpy
ΔS = entropy
by putting values,
ΔG = 27 KJ/mol - 100K(0.09 Kj/mol.K)
ΔG = 27 KJ/mol - 9 KJ/mol
ΔG = 18KJ/mol
Its durability, malleability, etc the basic properties of metals
Answer:
A chemical equation consists of the chemical formulas of the reactants (on the left) and the products (on the right). The two are separated by an arrow symbol (“→” usually read aloud as “yields”). ... The equation also identifies that all the compounds are in the gaseous state.
Explanation:
