a. 1,4332 g
b. 7.54~g
<h3>Further explanation</h3>
Given
Reaction
MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)
20 cm of 2.5 mol/dm^3 of MgCl2
20 cm of 2.5 g/dm^3 of MgCl2
Required
the mass of silver chloride - AgCl
Solution
a. mol MgCl2 :

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1
mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g
b. mol MgCl2 (MW=95.211 /mol):

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526
mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g
 
        
             
        
        
        
The mass of lime that  can  be produced  from  4.510 Kg of limestone  is calculated  as  below
calculate the moles  of CaCO3  used
that is  moles =mass/molar mass
convert  Kg  to g  = 4.510 x1000 =4510g
=  4510 / 100 =45.10 moles
CaCO3 = CaO  +O2
by use of mole ratio between CaCO3  to CaO  (1:1) the  moles of CaO  is also= 45.10 moles
mass of CaO = moles x molar  mass
45.10  x56 = 2525.6  g  of  CaO
        
                    
             
        
        
        
Answer:
5200 ppm
Explanation:
As per the definition, parts per million of a contaminant is a measure of the amount of mass of contaminant present per million amount of the solution. It is denoted by ppm.
Given in the question,
Water = 250 ml = 250 g
Lead = 1.30 g
So,
ppm of Lead =  =
 =  = 5200 ppm
 = 5200 ppm
So, as calculated above, there is 5200 ppm of lead present in 250 ml of water.
 
        
             
        
        
        
Umm I think it might be a b c in order?
        
             
        
        
        
Answer:
Explanation:
Percent composition by element
Element	Symbol	# of Atoms
Chlorine	Cl	2
Calcium	Ca	1
Oxygen	O	6