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Chemistry-

General

Easy

Question

Ethylmercaptan is prepared by the reaction of the following, followed by hydrolysis

$C_{2} H_{5} MgBr + {SO}_{2}$
$C_{2} H_{5} MgBr + S$
$C_{2} H_{5} MgBr + {CS}_{2}$
$C_{2} H_{5} MgBr + H_{2} S$

The correct answer is: $straight C subscript 2 straight H subscript 5 MgBr plus straight S$

$straight C subscript 2 straight H subscript 5 MgBr plus straight S not stretchy ⟶ for straight H subscript 3 straight O to the power of circled plus of straight C subscript 2 straight H subscript 5 minus SH$

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Chemistry-General

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parallel

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A unit vector $stack a with rightwards arrow on top$ in the plane of $b with rightwards arrow on top equals 2 i with ˆ on top plus j with ˆ on top & c with rightwards arrow on top equals i with ˆ on top minus j with ˆ on top plus k with ˆ on top$ is such that $a with rightwards arrow on top b with rightwards arrow on top equals a with rightwards arrow on top d with rightwards arrow on top$ where $d with rightwards arrow on top equals j with ˆ on top plus 2 k with ˆ on top$ is

A unit vector $stack a with rightwards arrow on top$ in the plane of $b with rightwards arrow on top equals 2 i with ˆ on top plus j with ˆ on top & c with rightwards arrow on top equals i with ˆ on top minus j with ˆ on top plus k with ˆ on top$ is such that $a with rightwards arrow on top b with rightwards arrow on top equals a with rightwards arrow on top d with rightwards arrow on top$ where $d with rightwards arrow on top equals j with ˆ on top plus 2 k with ˆ on top$ is

If $a with rightwards arrow on top to the power of blank equals i with ˆ on top plus j with ˆ on top comma b with rightwards arrow on top to the power of ´ equals i with ˆ on top plus j with ˆ on top plus 2 k with ˆ on top & c with rightwards arrow on top to the power of ´ equals 2 i plus j with ˆ on top minus k with ˆ on top$ . Then altitude of the parallel piped formed by the vectors $stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top$ having base formed by $stack b with rightwards arrow on top & stack c with rightwards arrow on top$ is $open parentheses stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top close$ and $a with rightwards arrow on top to the power of blank comma b with rightwards arrow on top to the power of blank$ , $stack c with rightwards arrow on top to the power of ´ end exponent$ are reciprocal system of vectors)

If $a with rightwards arrow on top to the power of blank equals i with ˆ on top plus j with ˆ on top comma b with rightwards arrow on top to the power of ´ equals i with ˆ on top plus j with ˆ on top plus 2 k with ˆ on top & c with rightwards arrow on top to the power of ´ equals 2 i plus j with ˆ on top minus k with ˆ on top$ . Then altitude of the parallel piped formed by the vectors $stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top$ having base formed by $stack b with rightwards arrow on top & stack c with rightwards arrow on top$ is $open parentheses stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top close$ and $a with rightwards arrow on top to the power of blank comma b with rightwards arrow on top to the power of blank$ , $stack c with rightwards arrow on top to the power of ´ end exponent$ are reciprocal system of vectors)

parallel

If the four faces of a tetrahedron are represented by the equations $r with minus on top times left parenthesis alpha i with minus on top plus beta j with minus on top right parenthesis equals 0 comma r with minus on top times left parenthesis beta j with minus on top plus gamma k with minus on top right parenthesis equals 0 comma r with minus on top times left parenthesis gamma k with minus on top plus alpha i with minus on top right parenthesis equals 0$ and $r with minus on top times left parenthesis alpha i with minus on top plus$ $beta j with minus on top plus gamma k with minus on top right parenthesis equals P$ then volume of the tetrahedron (in cubic units) is

If the four faces of a tetrahedron are represented by the equations $r with minus on top times left parenthesis alpha i with minus on top plus beta j with minus on top right parenthesis equals 0 comma r with minus on top times left parenthesis beta j with minus on top plus gamma k with minus on top right parenthesis equals 0 comma r with minus on top times left parenthesis gamma k with minus on top plus alpha i with minus on top right parenthesis equals 0$ and $r with minus on top times left parenthesis alpha i with minus on top plus$ $beta j with minus on top plus gamma k with minus on top right parenthesis equals P$ then volume of the tetrahedron (in cubic units) is

If $sum subscript i equals 1 end subscript superscript n end superscript stack a with rightwards arrow on top subscript i end subscript equals stack 0 with rightwards arrow on top$ where $open vertical bar stack a with rightwards arrow on top subscript i end subscript close vertical bar equals 1 for all i$ , then the value of $sum subscript text i end text less or equal than text end text j less or equal than n end subscript stack a subscript i with rightwards arrow on top times a with rightwards arrow on top subscript j$ is

If $sum subscript i equals 1 end subscript superscript n end superscript stack a with rightwards arrow on top subscript i end subscript equals stack 0 with rightwards arrow on top$ where $open vertical bar stack a with rightwards arrow on top subscript i end subscript close vertical bar equals 1 for all i$ , then the value of $sum subscript text i end text less or equal than text end text j less or equal than n end subscript stack a subscript i with rightwards arrow on top times a with rightwards arrow on top subscript j$ is

The vector $stack O A with minus on top equals negative 2 i with ˆ on top plus j with ˆ on top plus 2 k with ˆ on top$ turn by a right angle about origin $left parenthesis 0 right parenthesis$ so that it passes through $j with ˆ on top plus k with ˆ on top$ . then the vector in new position is

The vector $stack O A with minus on top equals negative 2 i with ˆ on top plus j with ˆ on top plus 2 k with ˆ on top$ turn by a right angle about origin $left parenthesis 0 right parenthesis$ so that it passes through $j with ˆ on top plus k with ˆ on top$ . then the vector in new position is

parallel

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