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 $2 H subscript 2 end subscript O subscript 2 end subscript left parenthesis l right parenthesis ⟶ 2 H subscript 2 end subscript O left parenthesis l right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis$

Chemistry-

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:

Chemistry-General

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

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

A system consists of a ball of mass M₂ and a uniform thin rod of mass M₁ and length d. The rod $text is end text omega$ attached to a frictionless horizontal table by a pivot at point P and initially rotates at an angular speed as shown in figure. The rod strikes the ball, which is initially at rest. As a result just after collision, the rod stops and ball moves in the direction shown. If collision is elastic, the ratio M₁ /M₂ is

In the system shown all the surfaces are frictionless while pulley and string are massless. Mass of block A is 2m and that of block B is m. Acceleration of block B immediately after system is released from rest is

In the arrangement shown, end A of light inextensible string is pulled with constant velocity v. The velocity of block B is

A right circular cone with semi-vertical angle $alpha$ rests on a rough incline plane is increased, the cone will slide before it topples over, ifqangle of inclination coefficient of friction

The Atwood machine shown is suspended from a spring balance. The mass on one hanger is M, that on the other is (M + m). Suppose the heavier side (right side) hanger is fastened to the top pulley by a thread. The scale reads (2M + m)g The thread is burned and the system accelerates. While the masses on the Atwood machine accelerates the spring balance reads.