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Gre4nikov [31]
3 years ago
15

When solid mg metal is put into an aqueous solution of snso4, solid sn metal and a solution of mgso4 result. write the net ionic

equation for the reaction?
Chemistry
2 answers:
kvv77 [185]3 years ago
5 0
Mg(s) + SnSO₄(aq) = MgSO₄(aq) + Sn(s)

Mg + Sn²⁺ + SO₄²⁻ = Mg²⁺ + SO₄²⁻ + Sn

Mg + Sn²⁺ = Mg²⁺ + Sn
mr_godi [17]3 years ago
4 0

Answer: Mg(s)+Sn^{2+}(aq)\rightarrow Mg^{2+}(aq)+Sn(s)

Explanation:

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The given chemical equation is:-

Mg(s)+SnSO_4(aq)\rightarrow MgSO_4(aq)+Sn(s)

The complete ionic equation is:

Mg(s)+Sn^{2+}(aq)+SO_4^{2-}(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)+Sn(s)

The SO_4^{2-} ions which are present on both the sides of the equation are not involved in net ionic equation.

Hence, the net ionic equation is:

Mg(s)+Sn^{2+}(aq)\rightarrow Mg^{2+}(aq)+Sn(s)

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A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
Degger [83]

Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


              = 58%



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