When sodium is dissolved in liquid ammonia,
Question: When sodium is dissolved in liquid ammonia, a solution of deep blue colour is obtained. The colour of the solution is due to (i) ammoniated electron (ii) sodium ion (iii) sodium amide (iv) ammoniated sodium ion Solution: Option (i)ammoniated electron is the answer....
Read More →In the synthesis of sodium carbonate,
Question: In the synthesis of sodium carbonate, the recovery of ammonia is done by treating NH4Cl with Ca(OH)2. The by-product obtained in this process is (i) CaCl2 (ii) NaCl (iii) NaOH (iv) NaHCO3 Solution: Option (i)CaCl2 is the answer....
Read More →Find the slope and the equation of the line passing through the points:
Question: Find the slope and the equation of the line passing through the points: (5, 3) and ( - 5, - 3) Solution: The slope of the equation can be calculated using $\mathrm{m}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}} \Rightarrow \frac{-3-3}{-5-5}=\frac{-6}{-10}$ $\mathrm{~m}=\frac{3}{5}$ Now using two point form of the equation of a line $\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)...
Read More →Amphoteric hydroxides react with both alkalies
Question: Amphoteric hydroxides react with both alkalies and acids. Which of the following Group 2 metal hydroxides is soluble in sodium hydroxide? (i) Be(OH)2 (ii) Mg(OH)2 (iii) Ca(OH)2 (iv) Ba(OH)2 Solution: Option (i)Be(OH)2 is the answer....
Read More →The solubility of metal halides depends on their nature,
Question: The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to (i) Ionic nature of lithium fluoride (ii) High lattice enthalpy (iii) High hydration enthalpy for lithium-ion. (iv) Low ionisation enthalpy of the lithium atom Solution: Option (ii)High lattice enthalpy is the answer....
Read More →The normal to the curve
Question: The normal to the curve $x^{2}=4 y$ passing through $(1,2)$ is A. $2 x+y=4$ B. $x-y=3$ C. $x+y=1$ D. $x-y=1$ Solution: Given that the curve $x^{2}=4 y$ Differentiating both the sides w.r.t. $x$, $4 \frac{d y}{d x}=2 x$ Slope of the tangent $\frac{d y}{d x}=\frac{1}{2} x$ For $(1,2)$ : $\Rightarrow \frac{d y}{d x}=\frac{1}{2}$ Equation of the normal: $\left(\mathrm{y}-\mathrm{y}_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(\mathrm{x}-\mathrm{x}_{1}\right)$ $\Rightarrow(y-2)=\fr...
Read More →The order of decreasing ionisation
Question: The order of decreasing ionisation enthalpy in alkali metals is (i) Na Li K Rb (ii) Rb Na K Li (iii) Li Na K Rb (iv) K Li Na Rb Solution: Option (iii)Li Na K Rb is the answer....
Read More →Some of the Group 2 metal halides are covalent
Question: Some of the Group 2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is (i) BeCl2 (ii) MgCl2 (iii) CaCl2 (iv) SrCl2 Solution: Option (i)BeCl2 is the answer....
Read More →Find the slope and the equation of the line passing through the points:
Question: Find the slope and the equation of the line passing through the points: ( - 1, 1) and (2, - 4) Solution: The slope of the equation can be calculated using $\mathrm{m}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}} \Rightarrow \frac{-4-1}{2-(-1)}=\frac{-5}{3}$ $\mathrm{~m}=-\frac{5}{3}$ Now using two point form of the equation of a line $\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right...
Read More →Metals form basic hydroxides.
Question: Metals form basic hydroxides. Which of the following metal hydroxide is the least basic? (i) Mg(OH)2 (ii) Ca(OH)2 (iii) Sr(OH)2 (iv) Ba(OH)2 Solution: Option (i)Mg(OH)2 is the answer...
Read More →The normal at the point
Question: The normal at the point $(1,1)$ on the curve $2 y+x^{2}=3$ is A. $x+y=0$ B. $x-y=0$ C. $x+y+1=0$ D. $x-y=1$ Solution: Given that the curve $2 y+x^{2}=3$ has a normal passing through point $(1,1)$. Differentiating both the sides w.r.t. $x$, $2 \frac{d y}{d x}+2 x=0$ Slope of the tangent $\frac{d y}{d x}=-x$ For $(1,1)$ : $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-1$ Equation of the normal: $\left(\mathrm{y}-\mathrm{y}_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(\mathrm{x}-\...
Read More →Which of the carbonates given
Question: Which of the carbonates given below is unstable in air and is kept in CO2 atmosphere to avoid decomposition. (i) BeCO3 (ii) MgCO3 (iii) CaCO3 (iv) BaCO3 Solution: Option (i)BeCO3 is the answer....
Read More →Metal carbonates decompose on heating
Question: Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates is most stable thermally? (i) MgCO3 (ii) CaCO3 (iii) SrCO3 (iv) BaCO3 Solution: Option (iv)BaCO3is the answer....
Read More →The normal at the point
Question: The normal at the point $(1,1)$ on the curve $2 y+x^{2}=3$ is A. $x+y=0$ B. $x-y=0$ C. $x+y+1=0$ D. $x-y=1$ Solution: Given that the curve $2 y+x^{2}=3$ has a normal passing through point $(1,1)$. Differentiating both the sides w.r.t. $x$, $2 \frac{d y}{d x}+2 x=0$ Slope of the tangent $\frac{d y}{d x}=-x$ For $(1,1)$ : $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-1$ Equation of the normal: $\left(\mathrm{y}-\mathrm{y}_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(\mathrm{x}-\...
Read More →Find the slope and the equation of the line passing through the points:
Question: Find the slope and the equation of the line passing through the points: (i) $(3,-2)$ and $(-5,-7)$ Solution: Slope of equation can be calculated using $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \Rightarrow \frac{-7-(-2)}{-5-3}=\frac{-5}{-8}$ $m=\frac{5}{8}$ Now using two point form of the equation of a line $y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$ where $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=$ slope of line $y-(-2)=\frac{5}{8}(x-3) \Rightarrow 8(y+2)=5(x-3)$ $8 y+16=5 x-15$ $5 x...
Read More →The reducing power of a metal depends on various factors.
Question: The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution. (i) Sublimation enthalpy (ii) Ionisation enthalpy (iii) Hydration enthalpy (iv) Electron-gain enthalpy Solution: Option (iii) Hydration enthalpyis the answer....
Read More →Alkali metals react with water vigorously
Question: Alkali metals react with water vigorously to form hydroxides and dihydrogen. Which of the following alkali metals reacts with water least vigorously? (i) Li (ii) Na (iii) K (iv) Cs Solution: Option (i)Li is the answer....
Read More →Find the equation of the line passing through the point P
Question: Find the equation of the line passing through the point P( - 3, 5) and perpendicular to the line passing through the points A(2, 5) and B( - 3, 6) Solution: As two points passing through line perpendicular to the line are given, we will calculate slope using two points. Let slopes of the two lines be m1 and m2. $\mathrm{m}_{1}=\frac{\mathrm{y}_{2-\mathrm{y}_{1}}}{\mathrm{x}_{2}-\mathrm{x}_{1}} \Rightarrow \frac{6-5}{-3-2}=-\frac{1}{5}$ $\mathrm{~m}_{1}=-\frac{1}{5}$ Now the slope of th...
Read More →The alkali metals are low melting.
Question: The alkali metals are low melting. Which of the following alkali metal is expected to melt if the room temperature rises to 30C? (i) Na (ii) K (iii) Rb (iv) Cs Solution: Option (iv) Csis the answer....
Read More →Find the equation of the line passing through the point
Question: Find the equation of the line passing through the point P(4, - 5) and parallel to the line joining the points A(3, 7) and B( - 2, 4). Solution: As two points passing through a line parallel to the line are given, we will calculate slope using two points(slope of parallel lines is equal). $\mathrm{m}=\frac{\mathrm{y}_{2-\mathrm{y}_{1}}}{\mathrm{x}_{2}-\mathrm{x}_{1}} \Rightarrow \frac{4-7}{-2-3}=\frac{-3}{-5}$ $\mathrm{m}=\frac{3}{5}$ Now using the slope - intercept form, we will find i...
Read More →Why does the water show a high boiling
Question: Why does the water show a high boiling point as compared to hydrogen sulphide? Give reasons for your answer. Solution: This is due to the formation of very stable intermolecular hydrogen bonding (bonding with other molecules of the same or different species) because of the presence of electronegative elements like oxygen, fluorine, nitrogen....
Read More →Why is water molecule polar?
Question: Why is water molecule polar? Solution: A polar molecule is a molecule that has a slight positive and a slight negative charge distribution. Due to the high electronegativity difference between oxygen and hydrogen atoms, polarity increases....
Read More →Find the equation of a line which cuts off intercept 5 on the x - axis and
Question: Find the equation of a line which cuts off intercept 5 on the x - axis and makes an angle of 600 with the positive direction of the x - axis. Solution: As intercept is given i.e. c = 5 and angle given so first we will find slope of line. $m=\tan \theta$ $m=\tan 60^{\circ} \Rightarrow \sqrt{3}$ Now using slope intercept form of the equation of a line $y=m x+c$ $y=(\sqrt{3}) x+5$ $(\sqrt{3}) x-y+5=0$ So, the required equation of line is $(\sqrt{3}) x-y+5=0$...
Read More →With the help of suitable examples,
Question: With the help of suitable examples, explain the property of H2O2 that is responsible for its bleaching action? Solution: H2O2 or hydrogen peroxide acts as a strong oxidizing agent both in acidic and basic media. It acts as a bleaching agent due to the release of nascent oxygen. H2O2 ⇌ H2O + O he nascent oxygen combines with colouring matter breaks the chemical bonds present, which in turn gets oxidised. Hence, the visible light is not absorbed and the cloth gets whitened....
Read More →An acidic solution of hydrogen peroxide behaves
Question: An acidic solution of hydrogen peroxide behaves as an oxidising as well as the reducing agent. Illustrate it with the help of a chemical equation. Solution: H2O2 + 2HI I2+ 2H2O Here H2O2 cat as an oxidizing agent HOCl + H2O2 H3O++ Cl+ O2 Here H2O2 act as reducing agent o, we can see that H2O2 acts as an oxidising agent in the first reaction as it increases the oxidation number of I- in HI from -1 to 0 in I2. Similarly, H2O2 acts as a reducing agent in the second reaction as it decrease...
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