ChemistryQ1–2 of 2 questions
Page 1 of 1 · English
1
ChemistryMediumMCQIIT JEE · 1977
Chemical equilibrium is dynamic in nature because
A
Equilibrium is maintained rapidly
B
The concentration of reactants and products become same at equilibrium
C
The concentration of reactants and products are constant but different
D
None
Solution
(C) Chemical equilibrium is considered dynamic because,at equilibrium,the rate of the forward reaction equals the rate of the backward reaction.
Even though the concentrations of reactants and products remain constant over time,the reactions do not stop; they continue to occur at equal rates in both directions.
Even though the concentrations of reactants and products remain constant over time,the reactions do not stop; they continue to occur at equal rates in both directions.
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2
ChemistryMCQIIT JEE · 1977
The coefficient of $x$ in the equation $x^2 + px + q = 0$ was taken as $17$ in place of $13$. Its roots were found to be $-2$ and $-15$. The roots of the original equation are:
A
$3, 10$
B
$-3, -10$
C
$-5, -8$
D
None of these
Solution
(B) The incorrect equation is $x^2 + 17x + q = 0$.
Given that the roots of this incorrect equation are $-2$ and $-15$,we can find $q$ using the product of roots:
$q = (-2) \times (-15) = 30$.
Now,substitute $q = 30$ into the original equation where the coefficient of $x$ was $13$:
$x^2 + 13x + 30 = 0$.
To find the roots of the original equation,factorize the quadratic:
$x^2 + 10x + 3x + 30 = 0$
$x(x + 10) + 3(x + 10) = 0$
$(x + 3)(x + 10) = 0$.
Thus,the roots are $x = -3$ and $x = -10$.
Given that the roots of this incorrect equation are $-2$ and $-15$,we can find $q$ using the product of roots:
$q = (-2) \times (-15) = 30$.
Now,substitute $q = 30$ into the original equation where the coefficient of $x$ was $13$:
$x^2 + 13x + 30 = 0$.
To find the roots of the original equation,factorize the quadratic:
$x^2 + 10x + 3x + 30 = 0$
$x(x + 10) + 3(x + 10) = 0$
$(x + 3)(x + 10) = 0$.
Thus,the roots are $x = -3$ and $x = -10$.
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