List of practice Questions

If two positive integers a and b are written as a = x3y2 and b = xy3, where x, y are prime numbers, then the result obtained by dividing the product of the positive integers by the LCM (a, b) is

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The nature of roots of the quadratic equation 9x2 – 6x – 2 = 0 is:

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The following equilibria are given : $N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3}\, K _{1}$ $N2 + O2 \rightleftharpoons 2NO\, K_2$ $H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2 O \, K_3$ The equilibrium constant of the reaction $2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2 O$ in terms of $K_1,\, K_2$ and $K_3$ is

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The neutralization of $NaOH$ by $H _{2} SO _{4}$ takes place as follows $H _{2} SO _{4}+2 NaOH \longrightarrow Na _{2} SO _{4}+ H _{2} O$

For complete neutralization Equivalents of acid = equivalents of base Equivalents of $NaOH =$ moles $\times$ acidity $=1 \times 1=1$ Equivalents of $H _{2} SO _{4}=\frac{x}{98} \times 2=\frac{x}{49}$ (Mol. mass of $H _{2} SO _{4}=98$ ) Putting the values $1 \times 1 =\frac{x}{49}$ $\Rightarrow x =49\, g$ but $H _{2} SO _{4}$ is $70 \%$ let $y g 70 \% H _{2} SO _{4}$ is required $\frac{70}{100} \times y =49$ $\Rightarrow y =70\, g$

The following equilibria are given : $N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3}\, K _{1}$ $N2 + O2 \rightleftharpoons 2NO\, K_2$ $H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2 O \, K_3$ The equilibrium constant of the reaction $2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2 O$ in terms of $K_1,\, K_2$ and $K_3$ is

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M in HTML stands for:

An element ‘M' has 50% of the electrons filled in the 3rd shell as in the 2nd shell. The atomic number of ‘M’ is:

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Find the missing one.

4	?	9
1	8	9
5	4	3

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