`0.05` moles of `NaHCO_(3)` will react with how many equivalent of `Mg(OH)_(2)`?A. `0.2` equivalentB. `0.05` equivalentC. `0.02` equivalentD. `0.01` equivalent

`0.05` moles of `NaHCO_(3)` will react with how many equivalent of `Mg

1.	`0.05` moles of `NaHCO_(3)` will react with how many equivalent of `Mg(OH)_(2)`?A. `0.2` equivalentB. `0.05` equivalentC. `0.02` equivalentD. `0.01` equivalent
Answer» Correct Answer - B Eq. of `NaHCO_(3)=` Eq. of `Mg(OH)_(2)` `0.05=` equation of `Mg(OH)_(2)` `:.` Equivalent of `Mg(OH)_(2)=0.05`

Discussion

No Comment Found

Related InterviewSolutions

`0.80 g` of sample of impure potassium dichromate was dissolved in water and made up to `500 mL` solution. `25 mL` of this solution treated with excess of `KI` in acidic medium and `I_(2)` liberated required `24 mL` of a sodium thiosulphate solution. `30 mL` of this sodium thiosulphate solution required `15 mL` of `N//20` solution of pure potassium dichromate. What was the percentage of `K_(2)Cr_(2)O_(7)` in given sample?A. `73.5%`B. `75.3%`C. `36.75%`D. none of these
`12. g` of an impure sample of arsenious oxide was dissolved in water containing `7.5 g` of sodium bicarbonate and the resulting solution was diluted to `250 mL`. `25 mL` of this solution was completely oxidised by `22.4 mL` of a solution of iodine. `25 mL` of this iodine solution reacted with same volume of a solution containing `24.8 g` of sodium thiosulphate `(Na_(2)S_(2)O_(3).5H_(2)O)` in one litre. Calculate teh percentage of arsenious oxide in the sample ( Atomic mass of `As=57`)
The unbalanced equation for the reaction of `P_(4)S_(3)` with nitrate in aqueous acidic medium is given below : `P_(4)S_(3)+NO_(3)^(-)rarrH_(3)PO_(4)+SO_(4)^(2-)+NO` the number of moles of water reuired per mole of `P_(4)S_(3)` is :A. 18B. `(8)/(3)`C. 8D. 28
Complete and balance the following in acidic medium: `Ag^(o+)+AsH_(3)rarrH_(3)AsO_(3)`
Balance the following equation stepwise: `Cr_(2)O_(7)^(2-) + Fe^(2+)++H^(o+)rarrCr^(3+) + Fe^(3+) + H_(2)O`A. `6`, `7`B. `6`, `14`C. `5`, `7`D. `5`, `14`
Balance the following equations: (i) `Cr_(2)O_(7)^(2-)+I^(-)+H^(+)rarrCr^(3+)+I_(2)+H_(2)O` (ii) `Ag^(+)+AsH_(3)rarrH_(3)AsO_(3)+H^(+)+....`
`Cr_(2)O_(7)^(-2)+I^(-)+H^(+)rarrCr^(+3)+I_(2)+H_(2)O` The equivalent weight of the reductant in the above equation is :- (At. wt. of Cr = 52, I = 127)A. 26B. 127C. 63.5D. 10.4
`Cr_(2)O_(7)^(2-)+Xoverset(H^(o+))rarrCr^(3+)+H_(2)O+ "oxidised product" of X, X` in the above reaction cannot beA. `C_(2)O_(4)^(2-)`B. `Fe^(2+)`C. `SO_(4)^(2-)`D. `SO_(2-)`
Which of the following is a set fo reducing agents ?A. ` Cr_2O_7^(2-), CrO_4^(2-), Na`B. `I^(-) , Na, Fe^(2+)`C. `F^(-) , Cl^(-) , MnO_4^(-)`D. `HNO_3, Fe^(2+), F_(2)`
`Cr_(2)O_(7)^(2-)overset(H^(+))rarrCr^(3+)`, Eq. wt. of `Cr_(2)O_(7)^(2-)` is :-A. mol. Wt/6B. mol. Wt/3C. mol. Wt/4D. mol. Wt/1

`0.05` moles of `NaHCO_(3)` will react with how many equivalent of `Mg(OH)_(2)`?A. `0.2` equivalentB. `0.05` equivalentC. `0.02` equivalentD. `0.01` equivalent

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment