MCQ
The silicates which contain $Si_2O_7^{-6}$ units are called
- ACyclic silicates
- BChain silicates
- ✓Pyro silicates
- DOrtho silicates
✓
Answer
Correct option: C.
Pyro silicates
c
$SiO_{3.5}^{ - 3}$ pyro sillicate
$SiO_{3.5}^{ - 3}$ pyro sillicate
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The compound($s$) which generate($s$) $\mathrm{N}_2$ gas upon thermal decomposition below $300^{\circ} \mathrm{C}$ is (are)
→$(A)$ $\mathrm{NH}_4 \mathrm{NO}_3$ $(B)$ $\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7$ $(C)$ $\mathrm{Ba}\left(\mathrm{N}_3\right)_2$ $(D)$ $\mathrm{Mg}_3 \mathrm{~N}_2$
The order of increasing reactivity towards $HCl$ of the following compounds will be
→$(1)$ $C{H_2} = C{H_2}$
$(2)$ ${(C{H_3})_2}C = C{H_2}$
$(3)$ $C{H_3}CH = CHC{H_3}$
For the reversible reaction in equilibrium
→${N_2}\left( g \right) + {O_2}\left( g \right)\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}2NO\left( g \right)$
$C_0 = Ce^{-2.1×10^{-3}\ t}$ for the forward reaction and
$C_0'= C'e^{-4.2×10^{-4}\ t}$ for the backward reaction, hence $K_c$ for the above equilibrium is
On adding $0.1\, M$ solution each of $[Ag^+], [Ba^{2+}], [Ca^{2+}]$ in a $Na_2SO_4$ solution, species first precipitated is $[K_{sp}\, BaSO_4 = 10^{-11}, K_{sp}\, CaSO_4 = 10^{-6}, K_{sp}\,Ag_2SO_4 = 10^{-5}]$
→Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately $6.023 \times 10^{23}$ ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
→A $4.0$ molar aqueous solution of $\mathrm{NaCl}$ is prepared and $500 \mathrm{~mL}$ of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: $\mathrm{Na}=23, \mathrm{Hg}=200 ; 1$ Faraday $=96500$ coulombs)
$1.$ The total number of moles of chlorine gas evolved is
$(A)$ $0.5$ $(B)$ $1.0$ $(C)$ $2.0$ $(D)$ $3.0$
$2.$ If the cathode is a $\mathrm{Hg}$ electrode, the maximum weight ( $\mathrm{g}$ ) of amalgam formed from this solution is
$(A)$ $200$ $(B)$ $225$ $(C)$ $400$ $(D)$ $446$
$3.$ The total charge (coulombs) required for complete electrolysis is
$(A)$ $24125$ $(B)$ $48250$ $(C)$ $96500$ $(D)$ $193000$
Give the answer question $1,2$ and $3.$
Given that $K_w$ for water is $10^{-13}\,M^2$ at $62\,^oC$ , compute the sum of $pOH$ and $pH$ for a neutral aqueous solution at $62\,^oC$
→The four quantum numbers of the valence electron of potassium are
(image) $\xrightarrow[Cold\,dil.]{KMn{{O}_{4}}/\overset{\Theta }{\mathop{O}}\,H}X\xrightarrow{HI{{O}_{4}}}Y\xrightarrow{\overset{\Theta }{\mathop{O}}\,H/\Delta }Z;\,Z$ is
→The density of $Cu$ is $8.94\,g\, cm^{-3}$. The quantity of electricity needed to plate an area $10\,cm\times10\,cm$ to a thickness of $10^{-2}\,cm$ using $CuSO_4$ solution would be, if atomic mass of $Cu$ is $63.5$ .................. $\mathrm{C}$
→Which of the following can show geometrical isomerism :-