Hydrogen Test

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Hydrogen Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

What is the condensation point of water?

A
0 degrees Celsius

B
0 degrees Fahrenheit

C
100 degrees Fahrenheit

D
212 degrees Fahrenheit

Solution

Condensation point is a temperature at which gas changes to liquid. Generally, boiling point is same as condensation point. For water boiling point is 100$$ ^{\circ}C $$, -212$$ ^{\circ}F $$. So, condensation point is also 212$$ ^{\circ}F $$.
Question 2
1 / -0

Boiling water reacts with _____________ gives their respective hydroxides and ammonia gas.

A
magnesium nitride

B
calcium nitride

C
aluminium nitride

D
all of the above

Solution

Boiling water reacts with magnesium nitride, calcium nitride, aluminium nitride gives their respective hydroxides and ammonia gas.
Question 3
1 / -0

Boiling water reacts with $$C_6H_5N^+_2Cl^-$$ to give

A
Aniline

B
Benzylamine

C
Phenol

D
Benzaldehyde

Solution
Question 4
1 / -0

Why is water a bent molecule and not linear?

A
Water has four electron domains repelling each other, causing the molecule to be bent.

B
There are not enough electrons involved for the molecule to be linear.

C
Because the oxygen in water can only make bent molecules.

D
The molecule alternates in between a bent geometry and a linear geometry.

Solution

VSEPR theory tells us water has a bent shape.
Water is a bent molecule rather than linear. The reason for this bent shape comes from the electronegative nature of the oxygen atom. The electronegativity of oxygen creates a dipole moment in the molecule
Explanation:
The central oxygen atom has four pairs of electrons. Two pairs are shared in single covalent bonds with hydrogen atoms. The other two pairs are not shared with any other atom (non-bonding pair).
Question 5
1 / -0

What is the volume of "20 volume $$H_2O_2$$" required to get $$5000\ cm^3$$ of oxygen at S.T.P?

A
$$250\ cm^3$$

B
$$50\ cm^3$$

C
$$100\ cm^3$$

D
$$125\ cm^3$$

Solution

$$\displaystyle 5000\ cm^3$$ of oxygen corresponds to 5 L.

20 volume $$\displaystyle H_2O_2$$ means that decomposition of 1 L of $$\displaystyle H_2O_2$$ gives 20 L of oxygen at STP.

5 L of oxygen at S.T.P will be obtained from decomposition of $$\displaystyle \dfrac {5}{20} \times 1 = 0.250 \: L \: H_2O_2$$.

This corresponds to $$\displaystyle 250 \: cm^3$$ of 20 volume $$\displaystyle H_2O_2$$.
Question 6
1 / -0

Choose the correct statement about the industrial preparation of hydrogen from water gas $$(CO+H_2)$$.

A
$$H_2$$ is removed through occlussion with $$Pd$$

B
$$CO$$ and $$H_2$$ are fractionally separated using differences in their densities

C
$$CO$$ is removed by absorption in $$ aq. Cu_2Cl_2$$ solution

D
$$CO$$ is oxidised to $$CO_2$$ with steam in the presence of a catalyst followed by absorption of $$CO_2$$ in alkali

Solution

$$CO$$ is first oxidised to $$CO_2$$ with steam in the presence of a catalyst followed by absorption of $$CO_2$$ in alkali.
Question 7
1 / -0

In transforming $$0.01\ mole$$ of $$PbS$$ to $$PbSO_{4}$$, the volume of $$10$$ volume $$H_{2}O_{2}$$ required will be:

A
$$11.2\ mL$$

B
$$22.4\ mL$$

C
$$33.6\ mL$$

D
$$44.8\ mL$$

Solution

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

$$0.04\ mol$$ of $$H_{2}O_{2}$$ is required to react with $$0.01\ mol$$ $$PbS$$.

Molarity of $$H_{2}O_{2}$$ may be calculated as,

Volume strength $$= Molarity \times 11.2$$

$$10 = M\times 11.2$$

$$M = 0.892$$

Number of moles $$= \dfrac {MV}{1000}$$

$$0.04 = \dfrac {0.892\times V}{1000}$$

$$V = 44.8\ mL$$.
Question 8
1 / -0

The standard potential of the reaction: $$H_2O+e^-\rightarrow\dfrac{1}{2}H_2+OH^-$$ at 298 K by using $$ K_w(H_2 O)=10^{-14}$$.

A
-0.828 V

B
0.828 V

C
2.0 V

D
-0.5 V

Solution

We know
$$\Delta G=-RT\ln K$$

$$\because \Delta G=-nFE$$

$$\therefore -nFE=-RT\ln K$$

$$\therefore -1\times 96500\times E=-8.314\times 298\ln (10^{-14})$$

$$\therefore E=-0.828$$ $$V$$
Question 9
1 / -0

Which of the following ions forms a hydroxide that is highly soluble in water?

A
$$K^{+}$$

B
$$Zn^{2+}$$

C
$$Ni^{2+}$$

D
$$Al^{3+}$$

Solution

1. $$K^+$$ ion will form most water soluble hydroxide because it is the most reactive of these metals and thus forms strong alkali $$KOH$$ which will undergo complete dissociation in water, thus making it highly soluble.
Question 10
1 / -0

Which of the following hydrolysis reaction is incorrect at moderate temperature?

A
$$4BF_{3} + 3H_{2}O \rightarrow H_{3}BO_{3} + 3HBF_{4}$$

B
$$2NF_{3} + 3H_{2}O\rightarrow N_{2}O_{3} + 6HF$$

C
$$NCl_{3} + 3H_{2}O\rightarrow NH_{3} + HOCl$$

D
$$BiCl_{3} + H_{2}O\rightarrow BiOCl\downarrow + 2HCl$$

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Hydrogen Test - 57

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Hydrogen Test - 57

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SHARING IS CARING

Question 1

0 degrees Celsius

0 degrees Fahrenheit

100 degrees Fahrenheit

212 degrees Fahrenheit

Solution

Question 2

magnesium nitride

calcium nitride

aluminium nitride

all of the above

Solution

Question 3

Aniline

Benzylamine

Phenol

Benzaldehyde

Solution

Question 4

Water has four electron domains repelling each other, causing the molecule to be bent.

There are not enough electrons involved for the molecule to be linear.

Because the oxygen in water can only make bent molecules.

The molecule alternates in between a bent geometry and a linear geometry.

Solution

Question 5

$$250\ cm^3$$

$$50\ cm^3$$

$$100\ cm^3$$

$$125\ cm^3$$

Solution

Question 6

$$H_2$$ is removed through occlussion with $$Pd$$

$$CO$$ and $$H_2$$ are fractionally separated using differences in their densities

$$CO$$ is removed by absorption in $$ aq. Cu_2Cl_2$$ solution

$$CO$$ is oxidised to $$CO_2$$ with steam in the presence of a catalyst followed by absorption of $$CO_2$$ in alkali

Solution

Question 7

$$11.2\ mL$$

$$22.4\ mL$$

$$33.6\ mL$$

$$44.8\ mL$$

Solution

Question 8

-0.828 V

0.828 V

2.0 V

-0.5 V

Solution

Question 9

$$K^{+}$$

$$Zn^{2+}$$

$$Ni^{2+}$$

$$Al^{3+}$$

Solution

Question 10

$$4BF_{3} + 3H_{2}O \rightarrow H_{3}BO_{3} + 3HBF_{4}$$

$$2NF_{3} + 3H_{2}O\rightarrow N_{2}O_{3} + 6HF$$

$$NCl_{3} + 3H_{2}O\rightarrow NH_{3} + HOCl$$

$$BiCl_{3} + H_{2}O\rightarrow BiOCl\downarrow + 2HCl$$

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