Chemistry-

General

Easy

### Question

#### Which step is rate determining ste

- Formation of II
- Formation of III
- Formation of V
- Formation of IV

#### The correct answer is: Formation of IV

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### Related Questions to study

chemistry-

#### Which reagent (X) is used to convert I to II

#### Which reagent (X) is used to convert I to II

chemistry-General

chemistry-

#### An organic compound A upon reacting with NH_{3} gives B. On heating B gives C. C in presence of KOH reacts with Br_{2} to given CH_{3}CH_{2}NH_{2} . A is :

#### An organic compound A upon reacting with NH_{3} gives B. On heating B gives C. C in presence of KOH reacts with Br_{2} to given CH_{3}CH_{2}NH_{2} . A is :

chemistry-General

Maths-

#### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is , then the number of ordered pair (p, q) is-

Step by step solution :

It is given that

That is, at least one of $p$ and $q$ must have $,$ and in their prime factorizations.

Now, consider the cases for power of $r$ as follows:

Case 1 : p contains then q has with K = (0,1,2,3)

That is, number of ways$=4$.

Case 2 : q contains then q has with K = (0,1,2,3)

That is, number of ways$=4$.

Case 3 : Both p and q contains

Then, number of ways$=1$.

Therefore, exponent of $r$ may be chosen in $2+2+1=5$ ways.

Similarly, exponent of s may be chosen in $4+4+1=9$ ways and

Exponent of $s$ may be chosen in $2+2+1=5$ ways

Thus, the total number of ways is:

Hence, the number of the ordered pair $(p,q)$ is $225$.

It is given that

That is, at least one of $p$ and $q$ must have $,$ and in their prime factorizations.

Now, consider the cases for power of $r$ as follows:

Case 1 : p contains then q has with K = (0,1,2,3)

That is, number of ways$=4$.

Case 2 : q contains then q has with K = (0,1,2,3)

That is, number of ways$=4$.

Case 3 : Both p and q contains

Then, number of ways$=1$.

Therefore, exponent of $r$ may be chosen in $2+2+1=5$ ways.

Similarly, exponent of s may be chosen in $4+4+1=9$ ways and

Exponent of $s$ may be chosen in $2+2+1=5$ ways

Thus, the total number of ways is:

Hence, the number of the ordered pair $(p,q)$ is $225$.

#### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is , then the number of ordered pair (p, q) is-

Maths-General

Step by step solution :

It is given that

That is, at least one of $p$ and $q$ must have $,$ and in their prime factorizations.

Now, consider the cases for power of $r$ as follows:

Case 1 : p contains then q has with K = (0,1,2,3)

That is, number of ways$=4$.

Case 2 : q contains then q has with K = (0,1,2,3)

That is, number of ways$=4$.

Case 3 : Both p and q contains

Then, number of ways$=1$.

Therefore, exponent of $r$ may be chosen in $2+2+1=5$ ways.

Similarly, exponent of s may be chosen in $4+4+1=9$ ways and

Exponent of $s$ may be chosen in $2+2+1=5$ ways

Thus, the total number of ways is:

Hence, the number of the ordered pair $(p,q)$ is $225$.

It is given that

That is, at least one of $p$ and $q$ must have $,$ and in their prime factorizations.

Now, consider the cases for power of $r$ as follows:

Case 1 : p contains then q has with K = (0,1,2,3)

That is, number of ways$=4$.

Case 2 : q contains then q has with K = (0,1,2,3)

That is, number of ways$=4$.

Case 3 : Both p and q contains

Then, number of ways$=1$.

Therefore, exponent of $r$ may be chosen in $2+2+1=5$ ways.

Similarly, exponent of s may be chosen in $4+4+1=9$ ways and

Exponent of $s$ may be chosen in $2+2+1=5$ ways

Thus, the total number of ways is:

Hence, the number of the ordered pair $(p,q)$ is $225$.

Maths-

#### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length –

We have, number of ways of selecting vertical sides = 1+3+5+....+(2m−1) which is equal to

Similarly, the number of ways of selecting horizontal sides = 1+3+5+....+(2n−1) which is equal to

So, the number of rectangles possible with odd side length =

number of ways of selecting vertical sides × number of ways of selecting horizontal sides.

Therefore, we get the number of rectangles possible with odd side length =

.

Similarly, the number of ways of selecting horizontal sides = 1+3+5+....+(2n−1) which is equal to

So, the number of rectangles possible with odd side length =

number of ways of selecting vertical sides × number of ways of selecting horizontal sides.

Therefore, we get the number of rectangles possible with odd side length =

#### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length –

Maths-General

We have, number of ways of selecting vertical sides = 1+3+5+....+(2m−1) which is equal to

Similarly, the number of ways of selecting horizontal sides = 1+3+5+....+(2n−1) which is equal to

So, the number of rectangles possible with odd side length =

number of ways of selecting vertical sides × number of ways of selecting horizontal sides.

Therefore, we get the number of rectangles possible with odd side length =

.

Similarly, the number of ways of selecting horizontal sides = 1+3+5+....+(2n−1) which is equal to

So, the number of rectangles possible with odd side length =

number of ways of selecting vertical sides × number of ways of selecting horizontal sides.

Therefore, we get the number of rectangles possible with odd side length =

chemistry-

#### Observe the esterification mechanisms for primary and tertiary alcohols

Type-1:

Type-2:

Mechanism:

In the above reaction (P) and (Q) are respectively :

#### Observe the esterification mechanisms for primary and tertiary alcohols

Type-1:

Type-2:

Mechanism:

In the above reaction (P) and (Q) are respectively :

chemistry-General

physics-

#### Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically work done is

#### Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically work done is

physics-General

physics-

#### A thermodynamic system is taken from state A to B along ACB and is brought back to Along BDA as Shown in the P$\to $V diagram. The net work done during the complete cycle is given by the area.

#### A thermodynamic system is taken from state A to B along ACB and is brought back to Along BDA as Shown in the P$\to $V diagram. The net work done during the complete cycle is given by the area.

physics-General

physics-

#### An ideal gas is taken V path ACBA as Shown in figure, The net work done in the whole cycle is

#### An ideal gas is taken V path ACBA as Shown in figure, The net work done in the whole cycle is

physics-General

physics-

#### In the Cyclic Process shown is the figure, the work done by the gas in one cycle

#### In the Cyclic Process shown is the figure, the work done by the gas in one cycle

physics-General

physics-

#### A Cyclic process is Shown in the P$\to $T diagram. Which of the curve show the same process on a V$\to $T diagram?

#### A Cyclic process is Shown in the P$\to $T diagram. Which of the curve show the same process on a V$\to $T diagram?

physics-General

physics-

#### The M.I of a disc of mass M and radius R about an axis passing through the centre O and perpendicular to the plane of dise is $\frac{{MR}^{2}}{2}$ If one quarter of the disc is removed the new moment of inertia of disc will be....

#### The M.I of a disc of mass M and radius R about an axis passing through the centre O and perpendicular to the plane of dise is $\frac{{MR}^{2}}{2}$ If one quarter of the disc is removed the new moment of inertia of disc will be....

physics-General

physics-

#### Work done in the given PV diagram in the cyclic process is

#### Work done in the given PV diagram in the cyclic process is

physics-General

physics-