Find the equation of a line which is equidistant from the lines
Question: Find the equation of a line which is equidistant from the lines x = - 2 and x = 6 Solution: For the equation of line equidistant from both lines, we will find point through which line passes and is equidistant from both line. As any point lying on $x=-2$ line is $(-2,0)$ and on $x=6$ is $(6,0)$, so mid - point is $(\mathrm{x}, \mathrm{y})=\left(\frac{-2+6}{2}, \frac{0+0}{2}\right)$ $(x, y)=(2,0)$ So, equation of line is x = 2....
Read More →Which of the following reactions increases
Question: Which of the following reactions increases production of dihydrogen from synthesis gas? (i) CH4 (g) + H2O (g) (1270K Ni)CO (g) + 3H2 (g) (ii) C (s) + H2O (g) (1270K) CO (g) + H2 (g) (iii) CO (g) + H2O (g) (Catalyst 673K) CO2 (g) + H2 (g) (iv) C2H6 + 2H2O (1270K Ni) 2CO + 5H2 Solution: Option (iii)CO (g) + H2O (g) (Catalyst 673K) CO2 (g) + H2 (g) is the answer....
Read More →Hydrogen peroxide is _________.
Question: Hydrogen peroxide is _________. (i) an oxidising agent (ii) a reducing agent (iii) both an oxidising and a reducing agent (iv) neither oxidising nor reducing agent Solution: Option (iii)both an oxidising and a reducing agent is the answer....
Read More →Find the equation of a vertical line passing through the point
Question: Find the equation of a vertical line passing through the point ( - 5, 6). Solution: Equation of line parallel to y - axis (vertical) is given by x = constant, as x - coordinate is constant for every point lying on line i.e. 6. So, the required equation of line is given as x = 6....
Read More →Which of the following equation depicts
Question: Which of the following equation depicts reducing nature of H2O2? (i) 2[Fe(CN)6]4 + 2H++ H2O2 2[Fe (CN)6]3 + 2H2O (ii) I2 + H2O2 + 2OH 2I + 2H2O + O2 (iii) Mn2+ + H2O2 Mn4+ + 2OH (iv) PbS + 4H2O2 PbSO4 + 4H2O Solution: Option (ii)I2 + H2O2 + 2OH 2I + 2H2O + O2 is the answer....
Read More →Which of the following equations depict
Question: Which of the following equations depict the oxidising nature of H2O2? (i) 2MnO4+ 6H++ 5H2O2 2Mn2++ 8H2O +5O2 (ii) 2Fe + 2H + H2O2 2Fe2++ 2H2O +O2 (iii) 2I+ 2H++ H2O2 I2+ 2H2O (iv) KIO4+ H2O2 KIO3+ H2O + O2 Solution: Option (iii)2I+ 2H++ H2O2 I2+ 2H2O is the answer....
Read More →The point on the curve
Question: The point on the curve $y=6 x-x^{2}$ at which the tangent to the curve is inclined at $\frac{\pi}{4}$ to the line $x+y=0$ is A. $(-3,-27)$ B. $(3,9)$ C. $\left(\frac{7}{2}, \frac{35}{4}\right)$ D. $(0,0)$ Solution: The curve $y=6 x-x^{2}$ has a point at which the tangent to the curve is inclined at to $\frac{\pi}{4}$ the line $x+y=0$. Differentiating w.r.t. $x$, $\frac{\mathrm{dy}}{\mathrm{dx}}=6-2 \mathrm{x}=\mathrm{m}_{1}$ and $\frac{\mathrm{dy}}{\mathrm{dx}}=-1=\mathrm{m}_{2}$ $\tan...
Read More →Find the equation of a horizontal line passing through the point
Question: Find the equation of a horizontal line passing through the point (4, - 2). Solution: Equation of line parallel to x - axis (horizontal) is y = constant, as y - coordinate of every point on the line parallel to $x$ - axis is $-2 i . e$. constant. Therefore equation of the line parallel to $x$ - axis and passing through $(4,-2)$ is $y=-2$....
Read More →The oxide that gives H2O2 on treatment
Question: The oxide that gives H2O2 on treatment with dilute H2SO4 is (i) PbO2 (ii) BaO2 .8H2O + O2 (iii) MnO2 (iv) TiO2 Solution: Option (ii) BaO2 .8H2O + O2is the answer....
Read More →Find the equation of a line parallel to the
Question: Find the equation of a line parallel to the x - axis and having intercept - 3 on the y - axis. Solution: Equation of line parallel to x - axis is given by y = constant, as x - coordinate of every point on the line parallel to y - axis is - 3 i.e. constant. So, the required equation of line is y = - 3....
Read More →Consider the reactions
Question: Consider the reactions (A) H2O2 + 2HI I2 + 2H2O (B) HOCl + H2O2 H3O++ Cl+ O2 Which of the following statements is correct about H2O2 with reference to these reactions? Hydrogen peroxide is ________. (i) an oxidising agent in both (A) and (B) (ii) an oxidising agent in (A) and reducing agent in (B) (iii) a reducing agent in (A) and oxidising agent in (B) (iv) a reducing agent in both (A) and (B) Solution: Option (ii)an oxidising agent in (A) and reducing agent in (B)is the answer....
Read More →Radioactive elements emit α, β
Question: Radioactive elements emit , and rays and are characterised by their halflives. The radioactive isotope of hydrogen is (i) Protium (ii) Deuterium (iii) Tritium (iv) Hydronium Solution: Option (iii) Tritium is the answer....
Read More →Find the equation of a line parallel to the y - axis at a distance of
Question: Find the equation of a line parallel to the y - axis at a distance of (i) 6 units to its right (ii) 3 units to its left Solution: (i) Equation of line parallel to y - axis is given by x = constant, as the x - coordinate of every point on the line parallel to $y$ - axis is 6 i.e. constant. Now the point lies to the right of $y$-axis means in the positive direction of $x$-axis, So, required equation of line is x = 6. (ii) Equation of line parallel to $y$-axis is given by $x=$ constant, a...
Read More →Which of the following hydrides
Question: Which of the following hydrides is electron-precise hydride? (i) B2H6 (ii) NH3 (iii) H2O (iv) CH4 Solution: Option (iv)CH4 is the answer....
Read More →Metal hydrides are ionic,
Question: Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH, CsH, the correct order of increasing ionic character is (i) LiH NaH CsH KHRbH (ii) LiH NaH KH RbH CsH (iii) RbH CsH NaH KH LiH (iv) NaH CsH RbH LiH KH Solution: Option (ii) is the answer....
Read More →Why does H+ ion always get associated
Question: Why does H+ ion always get associated with other atoms or molecules? (i) Ionisation enthalpy of hydrogen resembles that of alkali metals. (ii) Its reactivity is similar to halogens. (iii) It resembles both alkali metals and halogens. (iv) Loss of an electron from hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to the small size, it cannot exist free. Solution: Option (iv) Loss of an electron from hydrogen atom results in a nucleus of veryis...
Read More →Find the equation of a line parallel to the x - axis at a distance of
Question: Find the equation of a line parallel to the x - axis at a distance of (i) 4 units above it (ii) 5 units below it Solution: (i) Equation of line parallel to $x$-axis is given by $y=$ constant, as the $y$ - coordinate of every point on the line parallel to $x$ - axis is 4, i.e. constant. Now the point lies above $x$-axis means in positive direction of $y$-axis, So, the equation of line is given as y = 4. (ii) Equation of line parallel to $x$-axis is given by $y=$ constant, as the $y$ - c...
Read More →Hydrogen resembles halogens in many respects
Question: Hydrogen resembles halogens in many respects for which several factors are responsible. Of the following factors which one is most important in this respect? (i) Its tendency to lose an electron to form a cation. (ii) Its tendency to gain a single electron in its valence shell to attain stable electronic configuration. (iii) Its low negative electron gain enthalpy value. (iv) Its small size. Solution: Option (ii)Its tendency to gain a single electron in its valence shell to attain stab...
Read More →If the curves
Question: If the curves $y=2 e^{x}$ and $y=a e^{-x}$ interest orthogonally, then $a=$ A. $\frac{1}{2}$ B. $-\frac{1}{2}$ C. 2 D. $2 e^{2}$ Solution: Given that the curves $y=2 e^{x}$ and $y=a e^{-x}$ Differentiating both of them w.r.t. $x$, $\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{e}^{\mathrm{x}}$ and $\frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{ae}^{-\mathrm{x}}$ Let $\mathrm{m}_{1}=2 \mathrm{e}^{\mathrm{x}}$ and $\mathrm{m}_{2}=-\mathrm{ae}^{-\mathrm{x}}$ $m_{1} \times m_{2}=-1$ (Because curves...
Read More →A(1, 1), B(7, 3) and C(3, 6) are the vertices of a ΔABC
Question: A(1, 1), B(7, 3) and C(3, 6) are the vertices of a ΔABC. If D is the midpoint of BC and AL BC, find the slopes of (i) AD and (ii) AL. Solution: Given points are A(1, 1), B(7, 3) and C(3, 6) slope $=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$ Slope of line $B C=\left(\frac{3-6}{7-3}\right)=\frac{-3}{4}$ (i) As $D$ is the midpoint of $B C$, coordinate of $D$ are $D\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$ $=\left(\frac{7+3}{2}, \frac{3+6}{2}\right)=\left(5, \frac{9}{2}...
Read More →The equation of the normal to the curve
Question: The equation of the normal to the curve $x=\operatorname{acos}^{3} \theta, y=a \sin ^{3} \theta$ at the point $\theta=\frac{\pi}{4}$ is A. $\mathrm{x}=0$ B. $y=0$ C. $x=y$ D. $x+y=a$ Solution: Given that the curve $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ have a normal at the point $\theta=\frac{\pi}{4}$ Differentiating both w.r.t. $\theta$, $\frac{\mathrm{dx}}{\mathrm{d} \theta}=-3 \cos ^{2} \theta \sin \theta, \frac{\mathrm{dy}}{\mathrm{d} \theta}=3 \operatorname{asin}^{2} \theta ...
Read More →Which method can be used to find out
Question: Which method can be used to find out the strength of reductant/oxidant in a solution? Explain with an example. Solution: Strength of a reductant (reducing agent) or oxidant (oxidising agent) can be found out by measuring the relative electrode potential when its connected in a solution using a cell. For example, Fe3+/Fe is the element we want to test with the Standard Hydrogen electrode (SHE). The half-cell reaction for Fe and H are given below. H+ + e- H2 E = 0.0V Fe3+ + e- Fe2+ E = 0...
Read More →Show that the points A(0, 6), B(2, 1) and C(7, 3) are three corners
Question: Show that the points A(0, 6), B(2, 1) and C(7, 3) are three corners of a square ABCD. Find (i) the slope of the diagonal BD and (ii) the coordinates of the fourth vertex D. Solution: In a square, all sides are perpendicular to the adjacent side, so the product of slope of two adjacent sides is -1. Let the position of point $D(a, b)$. Given points of the square are $\mathrm{A}(0,6), \mathrm{B}(2,1), \mathrm{C}(7,3)$ and $\mathrm{D}(\mathrm{a}, \mathrm{b})$. slope $=\left(\frac{\mathrm{y...
Read More →Find out the oxidation number of chlorine
Question: Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine. NaClO4, NaClO3, NaClO, KClO2, Cl2O7, ClO3, Cl2O, NaCl, Cl2, ClO2 Solution: NaClO4 x= +7 NaClO3, x= +5 NaClO, x=+1 KClO2, x= +3 Cl2O7, x= +7 ClO3, x= +6 Cl2O, x=+1 NaCl, x=-1 Cl2, x=0 ClO2, x= +4 Ascending order of compounds w.r.t their oxidation number is: NaCl (-1), Cl2(0), Cl2O(+1), KClO2(+3), ClO2(+4), NaClO3(+5), ClO3(+6), Cl2O7=NaClO4(+7)....
Read More →The curves
Question: The curves $y=a e^{x}$ and $y=b e^{-x}$ cut orthogonally, if A. $a=b$ B. $a=-b$ C. $a b=1$ D. $a b=2$ Solution: Given that the curves $y=a e^{x}$ and $y=b e^{-x}$ Differentiating both of them w.r.t. $x$, $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{ae}^{\mathrm{x}}$ and $\frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{be}^{-\mathrm{x}}$ Let $m_{1}=a e^{x}$ and $m_{2}=-b e^{-x}$ $\mathrm{m}_{1} \times \mathrm{m}_{2}=-1$ (Because curves cut each other orthogonally) $\Rightarrow-a b=-1$ $\Rightarrow...
Read More →