The nitrogen containing compound produced in the reaction of HNO3 with P4O10
The correct statements regarding (i) HClO, (ii) HClO2, (iii) HClO3 and (iv) HClO4 is (are)
The correct statement(s) about O3 is(are)
With respect to graphite and diamond, which of the statement(s) given below is(are) correct?
Which of the following hydrogen halides reacts with AgNO3(aq.) to give a precipitate that dissolves in Na2S2O3 (aq.) ?
In the reaction
$\mathrm{2X+B_2H_6\to[BH_2(X)_2]^+[BH_4]^-}$
the amine(s) X is(are) :
The nitrogen oxide(s) that contain(s) N-N bond(s) is(are)
A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt(s) H is(are):
Dissolving $1.24 \mathrm{~g}$ of white phosphorous in boiling $\mathrm{NaOH}$ solution in an inert atmosphere gives a gas $\mathbf{Q}$. The amount of $\mathrm{CuSO}_{4}$ (in g) required to completely consume the gas $\mathbf{Q}$ is _________.
[Given: Atomic mass of $\mathrm{H}=1, \mathrm{O}=16, \mathrm{Na}=23, \mathrm{P}=31, \mathrm{~S}=32, \mathrm{Cu}=63$ ]
Explanation:
$\mathop {{P_4}}\limits_{\matrix{ {1.24\,g} \cr {or} \cr {0.01\,mole} \cr } } +3NaOH+3H_2O\to PH_3+3NaH_2PO_2$
As NaOH is present in excess. So, amount of phosphine formed is 0.01 mole (as P4 is limiting)
$\mathop {2P{H_3}}\limits_{0.01\,mole} +3CuSO_4\to Cu_3P_2+3H_2SO_4$
Amount of CuSO4 required = $\frac{3\times0.01}{2}$ mole
Mass of CuSO4 (in g) required = $\frac{0.03}{2}\times(63+32+16\times4)$
$=\frac{0.03}{2}\times159=2.38$ g
Explanation:
B2H6 and B3N3H6 have polar bond, but do not have same kind of atom.
Explanation:
$Xe{F_4} + {O_2}{F_2}\buildrel {143\,K} \over \longrightarrow \mathop {Xe{F_6}}\limits_{(Y)} + {O_2}$
The structure of XeF6 is
Y compound (XeF6) has 3 lone pair in each fluorine and one lone pair in xenon.
Hence, total number of lone pairs electrons is 19.
${N_2}{O_3},{N_2}{O_5},$ ${P_4}{O_6},{P_4}{O_7},$ ${H_4}{P_2}{O_5},{H_5}{P_3}{O_{10}},$ ${H_2}{S_2}{O_3},{H_2}{S_2}{O_5}$
Explanation:
The structures of various molecules given in problem are discussed below -
1. N2O3 It is the tautomeric mixture of following two structures -
Conclusion 1 bridging oxo group is present in the compound.
2. N2O5 It has following structure.
Conclusion 1 bridging oxo group is present in the compound.
where, pi = initial pressure, pt = total pressure, y = number of gaseous products per mole of reactant
3. P4O6
Conclusion 6 bridging oxo groups are present in the compound.
4. P4O7
Conclusion 6 bridging oxo groups are present in the compound.
5. H4P2O5
Conclusion 1 bridging oxo group is present in the compound.
6. H5P3O10
Conclusion 2 bridging oxo groups are present in the compound.
7. H2S2O3
Conclusion This compound does not contain any bridging oxo group.
8. H2S2O5
Conclusion This compound also does not contain any bridging oxo group.
Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing product formed is ____________.
Explanation:
The reaction is
B2H6 + 6CH3OH $\to$ 2B(OCH3)3 + 6H2
From the reaction, 1 mol of B2H6 reacts with 6 mol of CH3OH to produce 2 mol of B(OCH3)3.
Therefore, 3 mol of B2H6 would react with 18 mol of CH3OH to produce 6 mol of B(OCH3)3.
The number of lone pairs of electrons in N2O3 is ___________.
Explanation:
The structure of N2O3 is
Therefore, the total number of lone pairs is 8.
Among the following, the number of compounds that can react with PCl5 to given POCl3 is _____________.
O2, CO2, SO2, H2O, H2SO4, P4O10
Explanation:
The following reactions show how PCl5 reacts with different compounds to form POCl3 :
$CO_2 + PCl_5 \rightarrow POCl_3 + COCl_2$
$H_2O + PCl_5 \rightarrow POCl_3 + 2HCl$
$SO_2 + PCl_5 \rightarrow POCl_3 + SOCl_2$
$P_4O_{10} + 6PCl_5 \rightarrow 10POCl_3$
$H_2SO_4 + PCl_5 \rightarrow POCl_3 + HSO_3Cl + HCl$
From these reactions, it is clear that CO2, H2O, SO2, P4O10, and H2SO4 can react with PCl5 to give POCl3.
There are 5 compounds that meet this criterion.
The value of $n$ in the molecular formula $\mathrm{Be_n Al_2Si_6O_{18}}$ is ___________.
Explanation:
In the given molecular formula $\mathrm{Be_n Al_2Si_6O_{18}}$, according to charge balance in a molecule, we get
$2n+2(+3)+6(+4)-18(2)=0$
$\Rightarrow n=3$
So, the formula is $\mathrm{Be_3 Al_2Si_6O_{18}}$.
The total number of diprotic acids among the following is:
H3PO4, H2SO4, H3PO3, H2CO3, H2S2O7, H3BO3, H3PO2, H2CrO4 and H2SO3
Explanation:
A diprotic acid is an acid that contains within its molecular structure two hydrogen atoms per molecule capable of dissociating (i.e., ionisable protons) in water.
$ \mathrm{H}_2 \mathrm{SO}_4, \mathrm{H}_2 \mathrm{CO}_3, \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7, \mathrm{H}_2 \mathrm{CrO}_4, \mathrm{H}_3 \mathrm{PO}_3, \mathrm{H}_2 \mathrm{SO}_3 $
Structure of each of the compounds is given below :
(A) Calculate the amount of calcium oxide required to react with 852 g of P$_4$O$_{10}$.
(B) Write the structure of P$_4$O$_{10}$.
Explanation:
(A) The balanced chemical reaction between CaO and P$_4$O$_{10}$ is shown below.
$\mathrm{6CaO+P_4O_{10}\rightarrow 2Ca_3(PO_4)_2}$
Given, mass of $\mathrm{P_4O_{10}=852~g}$
The first step is to calculate the molar mass of $\mathrm{P_4O_{10}}$. The molar mass can be used to calculate the number of moles present in the given sample of $\mathrm{P_4O_{10}}$.
Molar mass of
$\mathrm{P_4O_{10}}=4\times31+10\times16=284$ g
Moles of $\mathrm{P_4O_{10}}=\frac{\mathrm{Mass~of~P_4O_{10}}}{\mathrm{Molar~mass~of~P_4O_{10}}}=\frac{852}{284}$
Moles of $\mathrm{P_4O_{10}}$ = 3 moles
1 mol of $\mathrm{P_4O_{10}}$ reacts with 6 mol of CaO to produce 2 mol of Ca$_3$(PO$_4$)$_2$.
$\therefore$ 3 mol of $\mathrm{P_4O_{10}}$ will react with 3 $\times$ 6 = 18 mol of CaO.
Mass of 1 mol of CaO = 40 + 16 = 56 g
$\therefore$ Mass of 18 mol of CaO = 18 $\times$ 56 = 1008 g
(B) Structure of $\mathrm{P_4O_{10}}$
In $\mathrm{P_4O_{10}}$, eqach P atom is bonded to 4 O atoms such that there are three P-O single bonds and one P-O double bond. Two phosphorous atoms share a single O atom.
Final Answer
(A) 1008 g
(B)