Hydrogen Test

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Hydrogen Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A person was using water supplied by municipality. Due to shortage of water, he started using underground water. He felt laxative effect. Its cause due to high concentration of:

A
$$CO_2$$

B
$$SO^{2-}_4$$

C
$$CO^{2-}_3$$

D
$$S^{2-}$$

Solution

The laxative effect is observed only when the sulphates present in water have concentration greater than $$500$$ppm. Otherwise at moderate levels, it is harmless.
Question 2
1 / -0

Pure water is natural because :

A
pH = 7

B
litmus has no effect

C
it is free from dissolved salts

D
pH=0

Solution

Pure water means water that has nothing in it except $$H_2O$$ (hydrogen and oxygen).

Water from various sources contains dissolved gases, minerals, organic substances.

So water is natural which is free from dissolved salts.

Hence the correct option is C.
Question 3
1 / -0

Water has maximum density at

A
Room temperature

B
$${4}^{o}C$$

C
$${0}^{o}C$$

D
$$-{4}^{o}C$$

Solution

When Water is in its Solid form (Ice) below $$0^0 C$$, it is forced into a regular Cage like Structure with small voids due to which the Volume of Ice is larger.

At $$4^0 C$$, the cage-like structure of ice starts to break and the molecules can slide along each other, closing the empty voids due to which momentarily the volume of water decreases. Due to decrease in volume, density increases because density is inversely proportional to volume.

As temperature rises to over $$4^0 C$$, the extra space needed by increased motion of water molecules starts being larger than the space gained due to structural changes and the molecules start to move away from each other due to which the volume again increases and density decreases.

Thus, Density maximum is reached at $$4^0C$$.
Question 4
1 / -0

Compound soluble in water which does not impede oxygen transportation is?

A
$$SO_2$$

B
$$SO_3$$

C
CO

D
NO

Solution

Oxygen transport is reduced in acidic conditions. $$SO_2$$ makes the blood less acidic as compared to $$SO_3$$. CO and NO combine with haemoglobin and reduce its oxygen carrying capacity.
Question 5
1 / -0

Hydrogen peroxide molecules are:

A
monoatomic and form $$X^{2-}_2$$ ions

B
diatomic and form $$X^-$$ ions

C
diatomic and form $$H{ O }^-_2$$ ions

D
monoatomic and form $$X^-$$ ions

Solution

$$H_2O_2$$ is a monoatomic molecule as it contains only one atom. It can be simply represented as $$HO-OH$$. It form $$(OH)_2^{2-}$$ ions on breakage. That is, it forms $$X_2^{2-}$$ ions.
$$HO-OH\rightleftharpoons {{(OH)}_2^{-2}}$$
Question 6
1 / -0

The volume strength of $$1$$ molar solution of $${ H }_{ 2 }{ O }_{ 2 }$$ is:

A
$$11.2$$

B
$$22.4$$

C
$$5.6$$

D
$$56$$

Solution

$$1$$ molar $${ H }_{ 2 }{ O }_{ 2 }$$ solution means $$1$$ mole (or $$34g$$) of ($${ H }_{ 2 }{ O }_{ 2 }$$) present in $${10}^{3}\,mL$$ solution.

$$\therefore$$ $$68g$$ $${ H }_{ 2 }{ O }_{ 2 }$$ gives $$=22400\,mL$$ of $${O}_{2}$$

$$\therefore$$ $$68g$$ $${ H }_{ 2 }{ O }_{ 2 }$$ gives $$=\cfrac { 22400 \,mL\times 34 }{ 68 } =11200\,mL$$ of $${O}_{2}$$

Volume strength $$=\cfrac { 11200 }{ { 10 }^{ 3 } } =11.2\ V$$
Question 7
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Products formed are:
$$2NaBH_4+I_2 \overset{Polyether}{\rightarrow}$$ ?

A
$$HI, Nal$$ and $$H_2$$

B
$$B_2H_6, Nal$$ and $$HI$$

C
$$B_2H_6, Nal$$ and $$H_2$$

D
$$H_3BO_3+H_2$$

Solution

$$NaBH_4+I_2\quad \underrightarrow {Polyether}\quad BH_3+HI+NaI$$
$$\downarrow$$ $$NaBH_4$$
$$B_2H_6+2NaI+H_2$$

Therefore, sodium borohydride reacts with iodine in presence of polyether to give diborane $$(B_2H_6)$$, sodium iodine and dihydrogen.

Answer: (C) $$B_2H_6,NaI$$ and $$H_2$$
Question 8
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Directions For Questions

A $$10\ ml$$ mixture of $$N_{2}$$, a alkane and $$O_{2}$$ undergo combustion in Eudiometry tube. There was contraction of $$2\ ml$$, when residual gases are passed through $$KOH$$. To the remaining mixture comprising of only one gas excess $$H_{2}$$ was added and after combustion the gas produced is absorbed by water, causing a reduction in volume of $$8\ ml$$.

...view full instructions

Gas produced after introduction of $$H_{2}$$ in the mixture?

A
$$H_{2}O$$

B
$$CH_{4}$$

C
$$CO_{2}$$

D
$$NH_{3}$$

Solution

Given
Volume of $$N_{2}$$, Alkane, $$O_{2}$$ is 10 ml
Solution
Reaction of $$N_{2}$$, $$C_{n}H_{2n+2}$$, $$O_{2}$$ gives $$CO_{2}$$
$$CO_{2}$$ is absorbed by KOH
Volume contracted = 2ml
Hence Volume of $$CO_{2}$$ = 2ml
Now in tube only $$N_{2}$$ and Alkane is left
When $$N_{2}$$ is reacted with excess of $$H_{2}$$ it forms $$NH_{3}$$.
The correct option is D
Question 9
1 / -0

In the Merck's process, the reagents involved in the preparation of hydrogen peroxide are:

A
$$BaO_{2}, HCl$$

B
$$Na_{2}O_{2}, H_{2}SO_{4}$$

C
$$BaO_{2}, H_{2}CO_{3}$$

D
$$PbO_{2}, H_{2}O$$

Solution

Merck's method is as follows:-

$$Na_2O_2+H_2SO_4\longrightarrow Na_2SO_4+H_2O_2$$

Sodium peroxide is added to $$20$$% ice cold solution of $$H_2SO_4$$ in small amounts with constant stirring.
On cooling, the $$Na_2SO_4.10H_2O$$ crystals separate out and the solution contains $$30$$% $$H_2O_2$$. A pure sample of $$H_2O_2$$ can be prepared by vacuum distillation.

Answer: (B) $$Na_2O_2, H_2SO_4$$
Question 10
1 / -0

Which reaction shows oxidising nature of $$H_{2}O_{2}$$?

A
$$Ag_{2}O + H_{2}O_{2} \rightarrow Ag + H_{2}O + O_{2}$$

B
$$MnO_{2} + H_{2} O_{2} + H_{2}SO_{4} \rightarrow MnSO_{4} + 2H_{2}O + O_{2}$$

C
$$PbS + 4H_{2}O_{2} \rightarrow PbSO_{4} + 4H_{2}O$$

D
$$K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 4H_{2}O_{2} \rightarrow K_{2}SO_{4} + 2CrO_{5} + 5H_{2}O$$

Solution

For (A), $${Ag}^{+1}_2O$$ is reduced to $$Ag^o$$
Therefore, $$H_2O_2$$ acts as a reducing agent.

For (B) $${Mn}^{+4}$$ is reduced to $${Mn}^{+2}$$
Here, $$H_2O_2$$ acts as a reducing agent.

For (C), There is no change in oxidation state of $$Pb$$.

For (D), $$Cr^{+6}$$ is oxidised to $$Cr^{+10}$$.
Therefore, $$H_2O_2$$ acts as an oxidising agent.

Answer: (D) $$K_2Cr_2O_7+H_2SO_4+4H_2O_2\longrightarrow K_2SO_4+2CrO_5+5H_2O$$

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Hydrogen Test - 41

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Hydrogen Test - 41

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SHARING IS CARING

Question 1

$$CO_2$$

$$SO^{2-}_4$$

$$CO^{2-}_3$$

$$S^{2-}$$

Solution

Question 2

pH = 7

litmus has no effect

it is free from dissolved salts

pH=0

Solution

Question 3

Room temperature

$${4}^{o}C$$

$${0}^{o}C$$

$$-{4}^{o}C$$

Solution

Question 4

$$SO_2$$

$$SO_3$$

CO

NO

Solution

Question 5

monoatomic and form $$X^{2-}_2$$ ions

diatomic and form $$X^-$$ ions

diatomic and form $$H{ O }^-_2$$ ions

monoatomic and form $$X^-$$ ions

Solution

Question 6

$$11.2$$

$$22.4$$

$$5.6$$

$$56$$

Solution

Question 7

$$HI, Nal$$ and $$H_2$$

$$B_2H_6, Nal$$ and $$HI$$

$$B_2H_6, Nal$$ and $$H_2$$

$$H_3BO_3+H_2$$

Solution

Question 8

$$H_{2}O$$

$$CH_{4}$$

$$CO_{2}$$

$$NH_{3}$$

Solution

Question 9

$$BaO_{2}, HCl$$

$$Na_{2}O_{2}, H_{2}SO_{4}$$

$$BaO_{2}, H_{2}CO_{3}$$

$$PbO_{2}, H_{2}O$$

Solution

Question 10

$$Ag_{2}O + H_{2}O_{2} \rightarrow Ag + H_{2}O + O_{2}$$

$$MnO_{2} + H_{2} O_{2} + H_{2}SO_{4} \rightarrow MnSO_{4} + 2H_{2}O + O_{2}$$

$$PbS + 4H_{2}O_{2} \rightarrow PbSO_{4} + 4H_{2}O$$

$$K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 4H_{2}O_{2} \rightarrow K_{2}SO_{4} + 2CrO_{5} + 5H_{2}O$$

Solution

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