MCQ
Using the rules for significant figures, the correct answer for the expression $\frac{0.02858 \times 0.112}{0.5702}$ will be .... .
- A$0.005613$
- ✓$0.00561$
- C$0.0056$
- D$0.006$
✓
Answer
Correct option: B.
$0.00561$
b
Reported answer should not be more precise than least precise term in calculations, so there should be three significant figures in reported answer.
Reported answer should not be more precise than least precise term in calculations, so there should be three significant figures in reported answer.
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 amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy ($BDE$) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below :$ Cl - Cl ( g ) \longrightarrow Cl ^*( g )+ Cl ^*( g ) \Delta H ^{\circ}=58 kcal mol ^{-1}$
$H _3 C - Cl ( g ) \longrightarrow H _3 C ^*( g )+ Cl ^{\circ}( g ) \Delta H ^{\circ}=85 kcal mol ^{-1}$
$H - Cl ( g ) \longrightarrow H ^*( g ) \quad+ Cl ^*( g ) \Delta H ^{\circ}=103 kcal mol ^{-1}$
$(1)$ Correct match of the $C - H$ bonds (shown in bold) in Column $J$ with their BDE in Column $K$ is
$(A) \ P - iii, Q - iv, R - ii, S - i$
$(B) \ P - i, Q - ii, R - iii, S - iv$
$(C) \ P - iii, Q - ii, R - i, S - iv$
$(D) \ P - ii, Q - i, R - iv, S - iii$
($2$) For the following reaction
$CH _4( g )+ Cl _2( g ) \xrightarrow{\text { light }} CH _3 Cl ( g )+ HCl ( g )$
the correct statement is
$(A)$ Initiation step is exothermic with $\Delta H ^{\circ}=-58 kcal mol ^{-1}$
$(B)$ Propagation step involving ${ }^{\circ} CH _3$ formation is exothermic with $\Delta H ^{\circ}=-2 kcal mol ^{-1}$.
$(C)$ Propagation step involving $CH _3 Cl$ formation is endothermic with $\Delta H ^{\circ}=+27 kcal mol ^{-1}$.
$(D)$ The reaction is exothermic with $\Delta H ^{\circ}=-25 kcal mol ^{-1}$.
→$H _3 C - Cl ( g ) \longrightarrow H _3 C ^*( g )+ Cl ^{\circ}( g ) \Delta H ^{\circ}=85 kcal mol ^{-1}$
$H - Cl ( g ) \longrightarrow H ^*( g ) \quad+ Cl ^*( g ) \Delta H ^{\circ}=103 kcal mol ^{-1}$
$(1)$ Correct match of the $C - H$ bonds (shown in bold) in Column $J$ with their BDE in Column $K$ is
| Column $J$ Molecule | Column $K$ $\operatorname{BDE}( kcal mol -1)$ |
| $(P) \ H - C H \left( CH _3\right)_2$ | ${ (i) } 132$ |
| $(Q) \ H - CH _2 Ph$ | ${ (ii) } 110$ |
| $(R) \ H - C H = CH _2$ | ${ (iii) } 95$ |
| $(S) \ H - C \equiv CH$ | ${ (iv) } 88$ |
$(B) \ P - i, Q - ii, R - iii, S - iv$
$(C) \ P - iii, Q - ii, R - i, S - iv$
$(D) \ P - ii, Q - i, R - iv, S - iii$
($2$) For the following reaction
$CH _4( g )+ Cl _2( g ) \xrightarrow{\text { light }} CH _3 Cl ( g )+ HCl ( g )$
the correct statement is
$(A)$ Initiation step is exothermic with $\Delta H ^{\circ}=-58 kcal mol ^{-1}$
$(B)$ Propagation step involving ${ }^{\circ} CH _3$ formation is exothermic with $\Delta H ^{\circ}=-2 kcal mol ^{-1}$.
$(C)$ Propagation step involving $CH _3 Cl$ formation is endothermic with $\Delta H ^{\circ}=+27 kcal mol ^{-1}$.
$(D)$ The reaction is exothermic with $\Delta H ^{\circ}=-25 kcal mol ^{-1}$.
If $0.5\, mol$ of $CaBr_2$ is mixed with $0.2\, mol$ of $K_3PO_4$ then the maximum number of moles of $Ca_3(PO_4)_2$ obtained will be
→A compound contains $2.8\%$ $Fe$ by mass. If the compound contains single $Fe$ atom in its one molecule, then the molar mass $(g/mol)$ of the compound is $(Fe = 56\, amu)$
→The following compound can exhibits
One mole of ${N_2}{H_4}$ loses $10\, mol$ of electrons to form a new compound $Y$. Assuming that all nitrogen appear in the new compound, what is the oxidation state of ${N_2}$ in $Y$ ? (There is no change in the oxidation state of hydrogen)
→Which of the following will not give $D.N.P.$ test ?
→Sulphur forms the chlorides ${S_2}C{l_2}{\rm{ \,and \,}}SC{l_2}$. The equivalent mass of sulphur in $SC{l_2}$ is......$g/mole$
→A metal $M$ which is not affected by strong acids like conc. $HNO_3,$ conc. $H_2SO_4$ and conc. solution of alkalies like $NaOH, KOH$ forms $MCl_3$ which finds use for toning in photography. The metal $M$ is
→In fire extinguisher, pyrene is
Complex compound $[Co(SCN)_2(NH_3)_4]Cl$ does not exhibit
→