std 9 gujarati ch.15 God matani kya ? swadhyay solution || dhoran 9 gujarati ch.15 swadhyay solution || ધોરણ 9 ગુજરાતી પાઠ 15 ગોદ માતાની ક્યાં? સ્વાધ્યાય સોલ્યુશન || std 9 gujarat board || std 9 gujarati

Algorithm to find the solution of quadratic equation by completing the square: 9x^(2)-15x+6=0

Show that the system of equations " " 6x + 5y = 11, 9x + ( 15) /(2) y = 21 . has no solution

Solution of the differential equation (9x+5y)dy+(15x+11y)dx=0 is

If it is given that x=9 is a solution of the equation ln(x^(2)+15a^(2))-ln(a-2)=ln((8ax)/(a-2)) then

Determine the pH of a 0.2 M solution of pyridine C_5H_5N. K_b= 1.5 xx 10^(-9)

Find the value of k for which following system of equutions have infinittly many solution: 4x+5y=3,quad kx+15y=9

Find the values of a and b for which the following system of equations has infinitely many solutions: 2x-(2a+5)y=5,quad (2b+1)x-9y=15

Objective question . i. A certains compound has the molecular formula X_(4) O_(6) . If 10 g of X_(4) O_(6) has 5.72 g X , then atomic mass of X is: a. 32 amu b. 42 amu c. 98 amu d. 37 amu ii. For 109% labelled oleum, if the number of moles of H_(2)SO_(4) and free SO_(3) be p and q , respectively, then what will be the value of (p - q)/(p + q) a. 1//9 b. 9 c. 18 d. 1//3 iii. Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation, 2H_(2) O_(2) (aq) rarr 2 H_(2) O (l) + O_(2) (g) Under conditions where 1 mol gas occupies 24 dm^(3), 100 cm^(3) of X M solution of H_(2) O_(2) produces 3 dm^(3) of O_(2) . Thus, X is a. 2.5 b. 0.5 c. 0.25 d. 1 iv. 4 g of sulphur is burnt to form SO_(2) which is oxidised by Cl_(2) water. The solution is then treated with BaCl_(2) solution. The amount of BaSO_(4) precipitated is: a. 0.24 mol b. 0.5 mol c. 1 mol d. 0.125 mol v. A reaction occurs between 3 moles of H_(2) and 1.5 moles of O_(2) to give some amount of H_(2) O . The limiting reagent in this reaction is a. H_(2) and O_(2) both b. O_(2) c. H_(2) d. Neither of them vi. 4 I^(ɵ) + Hg^(2+) rarr HgO_(4)^(-) , 1 mole each of Hg^(2+) and I^(ɵ) will form: a. 1 mol of HgI_(4)^(2-) b. 0.5 mol of HgI_(4)^(-2) 0.25 mol of HgI_(4)^(2-) 2 mol of HgI_(4)^(-2)