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
Answer:
The answer is D part of your lab notebook used for writing vocabulary
Hope it helps!
Use the question marck Moles of CO2
The the giving = 0.624 mol O2
Find the CF faction = 1 mole= 32.00 of O2
O= 2x16.00= 32.00amu ( writte this in the cf fraction)
SET UP THE CHART
Always start with the giving
0.624 mol O2 / 1mol of CO2
___________ / _____________ = Cancel the queal ( O2)
/ 32.00c O2
/
/
Multiply the top and divide by the bottom
0.624 mol CO x 1mol CO2 = 0.624 divide by 32.00 O2 =0.0195
You should look at the giving number ( how many num u gor ever there)
Ur answer should have the same # as ur givin so
= 0.0195
= .0195 mol of CO2
It will melt or the molecules inside of it will get hot
Here we go ~
1 mole of 
has 6.022 × 10²³ molecules of the given compound.
So, 0.78 mole of 
will have ~
