M = n/V
.5M = n/.100 L
n = .1 L * .5M
n= .05 mols of MgCl2
mass of MgCl2 = .05 mols of MgCl2 * 95.211 grams/ 1 mol of MgCl2 
mass of MgCl2 = 4.76 grams
4.76 grams of MgCl2 is needed to make 100 ml of a solution that is .500M, in chloride ion. Bolded = confused
        
             
        
        
        
Mutualism! The tree provides a home and food for the ants, while the ants protect the tree :)
        
             
        
        
        
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
Sodium borohydride is a relatively selective reducing agent  Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.
The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone 
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
to learn more about sodium borohydride and ethanol click here , 
brainly.com/question/12955502
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24.0mm^3=24÷10÷10÷10cm^3
density=Mass÷Volume