Oxidation states of the metal in the minerals haematite and magnetite, respectively, areA. II,III in haematite and III in magnetiteB. II,III in haemitite and II in magnetiteC. II in a haematite and II, III in magnetiteD. III in haematite and II, III in magnetite

Oxidation states of the metal in the minerals haematite and magnetite,

1.	Oxidation states of the metal in the minerals haematite and magnetite, respectively, areA. II,III in haematite and III in magnetiteB. II,III in haemitite and II in magnetiteC. II in a haematite and II, III in magnetiteD. III in haematite and II, III in magnetite
Answer» Correct Answer - D Haematite: `Fe_(2)O_(3):2a+3xx(-2)=0` `a=3`

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

Oxidation states of the metal in the minerals haematite and magnetite, respectively, areA. II,III in haematite and III in magnetiteB. II,III in haemitite and II in magnetiteC. II in a haematite and II, III in magnetiteD. III in haematite and II, III in magnetite

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment