Sodium carbonate solution reacts with strontium chloride solution. predict whether the reaction will produce a precipitate.

SrCO₃ will precipitate in the reaction

Explanation:

Na₂CO₃  +  SrCl₂   →  2 NaCl  +  SrCO₃ ↓

CO₃⁻² reacts with all the elements of group 2 of periodic table, to make an insoluble solid.

NaCl is a ionic salt, which is extremely soluble.

To solve this, you have to analyse all the reactions of cations and anions, with the qualitative inorganic analysis.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally