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#Science

The octahedral complex $\mathrm{CoSO_{4}Cl\cdot 5NH_{3}}$ exists in two isomeric forms $\mathrm{x}$ and $\mathrm{y}$ . Isomer $\mathrm{x}$ reacts with $\mathrm{AgNO_{3}}$ to give a while precipitate,but does not react with $\mathrm{BaCl_{2}}$ . Isomer $\mathrm{y}$ gives white precipitate with $\mathrm{BaCl_{2}}$ but does not react with $\mathrm{AgNO_{3}}$ .

Isomers $\mathrm{x}$ and $\mathrm{y}$ are
Option: 1
ionization isomers

Option: 2
linkage isomers

Option: 3
co-ordination isomers

Option: 4
solvate isomers

Ionization isomers have the same central metal ion; only the ligands exchange place with anion or neutral ion outside of the co-ordination complex. The sole difference between the two isomers is which ligands are attached to the central metal.

The X isomer of $\mathrm{CoS{{O}_{4}}Cl\cdot 5\text{ N}{{\text{H}}_{3}}}$ is $\mathrm{\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}S{{O}_{4}} \right]Cl}$ and the Y isomer is $\mathrm{\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}Cl \right]S{{O}_{4}}}$

Their reactions are as follows:

$\mathrm{\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}S{{O}_{4}} \right]Cl\text{ }\left( X \right)\text{ }\xrightarrow{AgN{{O}_{3}}}AgCl\downarrow \text{ (white ppt}\text{.)}}$

$\mathrm{\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}Cl \right]S{{O}_{4}}\text{ (Y) }\xrightarrow{BaC{{l}_{2}}}BaS{{O}_{4}}\downarrow \text{ (white ppt}\text{.)}}$

The correct answer is Option A- Ionization isomers.

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Posted by

Gautam harsolia

#Science

In a one component second order reaction,if the concentration of the reactant is reduced to half , the rate
Option: 1
increases two times

Option: 2
increases four times

Option: 3
decreases to one half

Option: 4
decreases to one fourth

As we have learnt,

For a second order Reaction

$\mathrm{Rate = k [Reactant]^2}$

So when the [Reactant] is halved, rate will become one-fourths

Hence, the correct answer is Option (4)

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Posted by

Gautam harsolia

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A massless string connects two pulley of masses ' $2 \mathrm{~kg}$ ' and ' $1 \mathrm{~kg}$ ' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
Option: 1
$\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2$

Option: 2
$\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2$

Option: 3
$\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2$

Option: 4
$\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2$

Not understanding sir

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Posted by

Raju vittal nandi

The following is the biochemical pathway for purple pigment production in flowers of sweet pea:

Colorless precursor 1 —- Allele A—-> Colorless precursor 2 —--- Allele B—---> Purple pigment

Recessive mutation of either gene A or B leads to the formation of white flowers. A cross is made between two parents with the genotype: AaBb × aabb. Considering that the two genes are not linked, the phenotypes of the expected progenies are:

Option: 1
9 purple : 7 white

Option: 2
3 white : 1 purple

Option: 3
1 purple : 1 white

Option: 4
9 purple : 6 light purple : 1 white

3(white):1(purple)

In case of 'duplicate recessive epistasis homozygous recessive gene masks the other gene i.e., aa is epistatic to A or a || bb is epistatic to B or b

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Posted by

Riya

JEE Main high-scoring chapters and topics

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Calculate the acceleration of block $m_1$ of the following diagram. Assume all surfaces are frictionless . Here m₁ = 100kg and m₂ = 50kg

Option: 1
0.33m/s²

Option: 2
0.66m/s²

Option: 3
1m/s²

Option: 4
1.32m/s²

1.32

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Posted by

DHFM DVP

No. of transition state in given figure

Option: 1
1

Option: 2
2

Option: 3
3

Option: 4
4

2

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Posted by

Mura

NEET 2024 Most scoring concepts

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When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?

Option: 1
G₁/S

Option: 2
G₂/M

Option: 3
M

Option: 4
Both G₂M and M

G₂/M should be activated as the cell has stalled DNA replication fork.

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Posted by

Ajit Kumar Dubey

A random variable X has the following probability distribution: $X: \: \: 1\; \; \: \: 2\; \;\: \: \: \: 3\; \; \: \: \: \: \: 4\: \: \: \; \; \: \: 5$ $P(X): \;K^{2}\; \; 2K\; \; K\; \; 2K\; \; 5K^{2}$ Then $P(X>2)$ is equal to:
Option: 1 $\frac{7}{12}$
Option: 2 $\frac{23}{36}$
Option: 3 $\frac{1}{36}$
Option: 4 $\frac{1}{6}$

Option 2

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Posted by

kpvalli2725@gmail.com

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Let $a_{n}$ be the n^th term of a G.P. of positive terms. If $\sum_{n=1}^{100}a_{2n+1}=200\: \: and\: \: \sum_{n=1}^{100}a_{2n}=100,\: \: then\: \sum_{n=1}^{200}a_n$ is equal to :
Option: 1 $300$
Option: 2 $175$
Option: 3 $225$
Option: 4 $150$

150

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Posted by

Nalla mahalakshmi

If $x=\sum_{n=0}^{\infty }(-1)^{n}\tan ^{2n}\theta \: \: and\: \: y=\sum_{n=0}^{\infty }\cos ^{2n}\theta ,$ for $0<\theta < \frac{\pi }{4},$ then
Option: 1 $y(1+x)=1$
Option: 2 $x(1-y)=1$
Option: 3 $y(1-x)=1$
Option: 4 $x(1+y)=1$

$x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta=1-\tan^2\theta+\tan^4\theta..........$

$y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta=1+\cos^2\theta+\cos^4\theta......$

Use $\text S_{\infty}=\frac{1}{1-r}$

${x=\frac{1}{1+\tan ^{2} \alpha}=\cos ^{2} \theta} \\ {y=\frac{1}{1-\cos ^{2} \theta}=\frac{1}{\sin ^{2} \theta}}$

$\Rightarrow (1-x)= \sin ^{2} \theta$

$\Rightarrow y(1-x)=1$

Correct Option (3)

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Posted by

avinash.dongre

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