The heats of combustion of rhombic and monoclinic sulphur are -70960 and -71030 calorie respectively What will be the heat of conversion of rhombic sulphur to monoclinic sulphur?

Heat of combustion of CH₄ , C₂, C₂H₄ and C₂H₄ are -212.8, -373.0, -337.0 and -310.5 kcal respectively at the same temperature. The best fuel among these gases is:

Standard heat of formation for CCI₄, H₂0, CO₂ and HCI at 298K are - 25.5, - 57.8, 94.1 and 22.1 kJl mol respectively. For the reaction,
$C C l subscript 4 end subscript plus 2 H subscript 2 end subscript O ⟶ C O subscript 2 end subscript plus 4 H C l$ what will be $capital delta H$

If $capital delta H subscript f end subscript superscript ring operator end superscript$ for $H subscript 2 end subscript O subscript 2 end subscript left parenthesis l right parenthesis$ and $H subscript 2 end subscript O left parenthesis l right parenthesis subscript negative 188 end subscript$ -188 kJ mol-I and-286kJ mol , what will be the enthalpy change of the reaction $2 H subscript 2 end subscript O subscript 2 end subscript left parenthesis l right parenthesis ⟶ 2 H subscript 2 end subscript O left parenthesis l right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis$

Which of the following units represents the largest amount of energy?

If $S plus O subscript 2 end subscript ⟶ S O subscript 2 end subscript semicolon capital delta H equals negative 298.2$ KJ
$S O subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction O subscript 2 end subscript ⟶ S O subscript 3 end subscript semicolon capital delta H equals negative 98.7$ KJ
$S O subscript 3 end subscript plus H subscript 2 end subscript O ⟶ H subscript 2 end subscript S O subscript 4 end subscript semicolon capital delta H equals negative 130.2$ KJ
$H subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction O subscript 2 end subscript ⟶ H subscript 2 end subscript O semicolon capital delta H equals negative 227.3$ KJ
the heat of foundation $o f blank H subscript 2 end subscript S O subscript 4 end subscript$ will be:

ABCD is a parallelogram and P is the intersection of the diagonals. If O is any point then OA+OB+OC+OD =

If the vectors $4 i with not stretchy bar on top minus 7 j with not stretchy bar on top minus 2 k with not stretchy bar on top comma i with not stretchy bar on top plus 5 j with not stretchy bar on top minus 3 k with not stretchy bar on top$ and $3 i with not stretchy bar on top minus lambda j with not stretchy bar on top plus k with not stretchy bar on top$ form a triangle then λ=

Let $f left parenthesis t with not stretchy bar on top right parenthesis equals left square bracket t right square bracket i with not stretchy bar on top minus left parenthesis t minus left square bracket t right square bracket right parenthesis j with not stretchy bar on top plus left square bracket t plus 1 right square bracket k with not stretchy bar on top comma left square bracket. right square bracket$ is greatest integer function. If $f open parentheses 5 over 4 close parentheses$ and $i with not stretchy bar on top plus lambda j with not stretchy bar on top plus mu k with not stretchy bar on top$ are parallel vectors then $left parenthesis lambda comma mu right parenthesis$ =

For such questions, we should know the concept of greatest integer number. We should also know the properties of parallel vectors.

Let $f left parenthesis t with not stretchy bar on top right parenthesis equals left square bracket t right square bracket i with not stretchy bar on top minus left parenthesis t minus left square bracket t right square bracket right parenthesis j with not stretchy bar on top plus left square bracket t plus 1 right square bracket k with not stretchy bar on top comma left square bracket. right square bracket$ is greatest integer function. If $f open parentheses 5 over 4 close parentheses$ and $i with not stretchy bar on top plus lambda j with not stretchy bar on top plus mu k with not stretchy bar on top$ are parallel vectors then $left parenthesis lambda comma mu right parenthesis$ =

For such questions, we should know the concept of greatest integer number. We should also know the properties of parallel vectors.

The points with position vectors $bar a plus bar b comma space bar a minus bar b$ and $a with not stretchy bar on top plus lambda b with not stretchy bar on top$ are collinear for

For such questions, we should know how to find the vector joining two points. Also, we should know the condition for two vectors to be collinear.

The points with position vectors $bar a plus bar b comma space bar a minus bar b$ and $a with not stretchy bar on top plus lambda b with not stretchy bar on top$ are collinear for

For such questions, we should know how to find the vector joining two points. Also, we should know the condition for two vectors to be collinear.

A= (1,1,1) B=(1,2,3) C=(2,-1,1) then the length of the internal bisector of $angle A$ is

A vector a has component $a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript$ in a right-handed rectangular Cartesian coordinate system OXYZ. The coordinate system is rotated about y-axis through an angle $fraction numerator text π end text over denominator text 2 end text end fraction$ . The components of a in the new system are

If $l$ ,m,n are the d.c’s of a vector if $l$ = $fraction numerator text 1 end text over denominator text 2 end text end fraction$ , then the maximum value of $l$ mn is

Physics-

A body performs simple harmonic oscillations along the straight line ABCDE with C as the midpoint of AE. Its kinetic energies at B and D are each one fourth of its maximum value. If AE = 2R, the distance between B and D is

Physics-General