Boiling of hard water is helpful in removing the temporary hardness by converting calcium hydrogen carbonate and magnesium hydrogen carbonate to ....
- A$CaCO_3$ and $Mg(OH)_2$
- B$CaCO_3$ and $MgCO_3$
- C$Ca(OH)_2$ and $MgCO_3$
- D$Ca(OH)_2$ and $Mg(OH)_2$
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Similar Questions
What is $Calgon$,used as a water softener?
Medium
View SolutionMatch the following:
List-$I$ (Reactions) List-$II$ (Methods) $A$. $Mg(HCO_3)_2 \rightarrow Mg(OH)_2 \downarrow + 2 CO_2 \uparrow$ $I$. Clark's method $B$. $M^{2+} + Na_4P_6O_{18}^{2-} \rightarrow [Na_2MP_6O_{18}]^{2-} + 2 Na^{+}$ $II$. Ion exchange method $C$. $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2 CaCO_3 + 2 H_2O$ $III$. Boiling $D$. $2 NaZ + Ca^{2+}_{(aq)} \rightarrow 2 Na^{+} + CaZ_2$ ($Z$ = Zeolite) $IV$. Calgon's method
| List-$I$ (Reactions) | List-$II$ (Methods) |
|---|---|
| $A$. $Mg(HCO_3)_2 \rightarrow Mg(OH)_2 \downarrow + 2 CO_2 \uparrow$ | $I$. Clark's method |
| $B$. $M^{2+} + Na_4P_6O_{18}^{2-} \rightarrow [Na_2MP_6O_{18}]^{2-} + 2 Na^{+}$ | $II$. Ion exchange method |
| $C$. $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2 CaCO_3 + 2 H_2O$ | $III$. Boiling |
| $D$. $2 NaZ + Ca^{2+}_{(aq)} \rightarrow 2 Na^{+} + CaZ_2$ ($Z$ = Zeolite) | $IV$. Calgon's method |
$A$ sample of water contains $Mg(HCO_3)_2$ and $Ca(HCO_3)_2$. On boiling this water, these hydrogen carbonates are removed as precipitates. The precipitates are
The temporary hardness of a water sample is due to compound $X$. Boiling this sample converts $X$ to compound $Y$. $X$ and $Y$,respectively,are
DifficultJEE MAIN 2019
View Solution$A$ sample of water contains $30 \ ppm$ of $MgSO_4$ and $19 \ ppm$ of $MgCl_2$. Calculate the hardness of water in $ppm$ (in terms of $CaCO_3$ equivalents).
Difficult
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