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Mr NaOH= (Ar Na )+(ArO)+(ArH) = 23+16+1=40
b. mol = 1.2/24 =0.05 mol
c. mol= 5.6/56=0.1 mol
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n= gr/mr
= 5/(23+16+1)
= 5/40
= 0.125
B. 0,5
n= gr/Ar
n= 1,2/24
= 0,5
C. 0,1
n= gr/Ar
n= 5,6/56