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Chemistry-

General

Easy

Question

The correct order of acidic strength is:

$A l_{2} O_{3} < S i O_{2} < P_{2} O_{3} < S O_{2}$
$S i O_{2} < S O_{2} < A l_{2} O_{3} < P_{2} O_{3}$
$A l_{2} O_{3} < S i O_{2} < S O_{2} < P_{2} O_{3}$
$S O_{2} < P_{2} O_{3} < S i O_{2} < A l_{2} O_{3}$

The correct answer is: $A l subscript 2 O subscript 3 less than S i O subscript 2 less than P subscript 2 O subscript 3 less than S O subscript 2$

Acidic character of oxides increases along the period.

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Let E ( $alpha$ ) = $open square brackets table row cell cos to the power of 2 end exponent invisible function application alpha end cell cell cos invisible function application alpha sin invisible function application alpha end cell row cell cos invisible function application alpha sin invisible function application alpha end cell cell sin to the power of 2 end exponent invisible function application alpha end cell end table close square brackets$ . If $alpha$ and $beta$ differs by an odd multiple of $pi$ /2,then E( $alpha$ ) E( $beta$ ) is a -

Let E ( $alpha$ ) = $open square brackets table row cell cos to the power of 2 end exponent invisible function application alpha end cell cell cos invisible function application alpha sin invisible function application alpha end cell row cell cos invisible function application alpha sin invisible function application alpha end cell cell sin to the power of 2 end exponent invisible function application alpha end cell end table close square brackets$ . If $alpha$ and $beta$ differs by an odd multiple of $pi$ /2,then E( $alpha$ ) E( $beta$ ) is a -

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If A = $open square brackets table row 0 cell negative 1 end cell 2 row 1 0 3 row cell negative 2 end cell cell negative 3 end cell 0 end table close square brackets$ , then A + 2A^T equals -

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For a general reaction given, below the value of solubility product can be given as :

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chemistry-General

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chemistry-General

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Assertion : The expression n!(20 – n)! is minimum when n = 10.
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an invertible matrix gives identity matrix as the product when multiplied with its inverse.

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the roots of the characteristic equation give us the eigenvalues of the matrix.

Let A = $open square brackets table row 1 0 0 row 0 1 1 row 0 cell negative 2 end cell 4 end table close square brackets$ & I = $open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets$ and A^-1= $fraction numerator 1 over denominator 6 end fraction$ [A²+CA+dI], find ordered pair (c,d) ?

the roots of the characteristic equation give us the eigenvalues of the matrix.

parallel

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