Chemistry Test

Result

Chemistry Test - 19

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A redox reaction is shown in the diagrams. Identify the reaction.

A
\(Z n_{(s)}+C u_{(a q)}^{2+} \rightarrow Z n_{(a q)}^{2+}+C u_{( s)}\)

B

\(C u_{(s)}+2 A g_{(a q)}^{+} \rightarrow C u_{(a q)}^{2+}+2 A g_{(s)}\)

C

\(2 A g_{\left(s\right)}+C u_{(a q)}^{2+} \rightarrow 2 A g_{(a q)}^{+}+C u_{( s)}\)

D

\(C u_{\left(s\right)}+Z n_{(a q)}^{2+} \rightarrow C u_{(a q)}^{2+}+Z n_{( s)}\)

Solution

The color changes of the solution and the metal rod indicate that the metal rod is of Zinc metal and the solution is of copper sulfate. When Zinc metal reacts with copper sulfate, Zinc metal gets oxidized and Copper sulfate gets reduced. The brown deposition is of the copper metal. The formed zinc sulfate is white in color and therefore, the solution becomes colorless.
Question 2
1 / -0

Which of the following is not a class of organic compounds?

A
Amides

B
Electrovalentcompounds

C
Nitro compound

D
Carbonyl compound

Solution

Classes of organic compounds are those which involves organic compounds such as carbon, hydrogen and oxygen. Hence, electrovalentcompounds is not a class of organic compounds.
Electrovalent compounds are formed by complete transfer of electrons. Electrovalent compounds are more soluble in polar solvents like water. An electrovalent compounds is formed when a metal atom transfers one or more electrons to a non-metal atom.
Question 3
1 / -0

For the reversible reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) 500^{\circ} \mathrm{C},\) the value of \(\mathrm{Kp}\) is \(1.44 \times 10^{-5}\) when partial pressure is measured in atmosphere. The corresponding value of \(\mathrm{K}_{\mathrm{c}},\) with concentration in mole litre \(^{-1}\), is -

A

\(\frac{1.44 \times 10^{-7}}{[(0.082 \times 773)^{-2}]}\)

B

\(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\)

C

\(\frac{1.44 \times 10^{-5}}{[(8.314 \times 773)^{-2}]}\)

D

\(\frac{1.44 \times 10^{-5}}{[(0.082\times 500)^{-2}]}\)

Solution

\(\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{r}}\left\{\begin{array}{l}\mathrm{n}_{\mathrm{p}}=\text { number of moles of products (gaseous) } \\ \mathrm{n}_{r}=\text { number of moles of reactants (gaseous) } \\ \qquad \Delta \mathrm{n}_{\mathrm{g}}=2-4=-2\end{array}\right.\)
\(\mathrm{K}_{\mathrm{C}}=\frac{\mathrm{K}_{\mathrm{P}}}{(\mathrm{RT})^{\Delta n_{\mathrm{g}}}}\)
\(\mathrm{Kc}=\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\left(\mathrm{R}\right.\) in \(\mathrm{L}\) atm mole \(\left.{ }^{-1} \mathrm{~K}^{-1}\right)\)
Question 4
1 / -0

Following tetrapeptide can be represented as

(F, L, D, Y, I, Q, P are one letter codes for amino acids)

A
FIQY

B
FLDY

C
YQLF

D
PLDY

Solution

Hydrolysis of the given tetrapeptide will give the following:
Question 5
1 / -0

Arrange the following conformational isomers of n -butane in order of their increasing potential energy

A
II < III

B
I < IV < III < II

C
II < IV < III < I

D
I < III < IV < II

Solution

The order of potential energy of above conformations is

The fully eclipsed form is least stable due to repulsion between bulky \(\left(-\mathrm{CH}_3\right)\) methyl group at front and rear carbon atom.

\(\therefore\) It has maximum potential energy.

While the repulsion in anti form is minimum.

\(\therefore\) It has minimum potential energy.
Question 6
1 / -0

When primary amines are treated with acid chlorides, we get:

A
Amides

B
Isocyanides

C
Acetaldehyde

D
Tertiary Amine

Solution

Acylation of amines:

Primary and secondary amines containing replaceable hydrogen atoms react with acid chlorides and acid anhydrides to form substituted amides.

Tertiary amines do not contain replaceable hydrogen atoms so these do not undergo reaction with an acid chloride or acid anhydride. The acid generated during the reaction can protonate the amine and destroy its nucleophilic character. Hence, a base is used. Unlike in the case of alkylation reaction of amines, the amide formed does not react further with organic halide or acid anhydride because the amide is non-basic and poor nucleophile due to the electron-withdrawing effect of -CO-R group. Acyl chlorides are stronger acylating agents than anhydrides and esters. So, when primary amines are treated with acid chlorides, we get amides.
Question 7
1 / -0

Which one of the following gases produces a red light when electricity flows through it?

A
Helium

B
Argon

C
Nitrogen

D
Neon
Solution
Neonproduces a red light when electricity flows through it.
- Electric discharge in gases occurs when electric current flows through a gaseous medium due to the ionization of the gas.
- Depending on several factors, the discharge may radiate visible light.
- Different elements emit different wavelengths of light to return to their respective ground states, so the colours of the tubes are varied.
- These colours can be used to produce atomic emission spectra of the elements electrically excited.
- Using known values of emission spectra, one can perform a similar discharge test on an unknown gas, gather an emission spectrum from it, and determine which elements are in the unknown gas.
Question 8
1 / -0

The total number of electrons that take part in forming bonds in \(\mathrm{N}_{2}\) is:

A
\(2\)

B
\(4\)

C
\(6\)

D
\(10\)

Solution

The total number of electrons that take part in forming bonds in \(\mathrm{N}_{2}\) is \(6 .\)

Structure of \(N_{2}\) is \(N \equiv N\).

We know that the atomic number of nitrogen is 7 and the electronic configuration of nitrogen is 2,5 which means two electrons are present in s-orbital and five electrons in \(\mathrm{p}-\) orbital.

When two nitrogen atoms will react, three electrons (from five valence electrons) of each nitrogen atom are shared between them so that both will obtain a stable configuration.

In \(N_{2}\) there are 3 bonds between two nitrogen atoms and each bond is formed by 2 electrons.

The full outer shells with the shared electrons are now stable forming a covalent bond and this can be shown through the following diagram:

So, 3 bonds make it \(3 \times 2=6\) electrons in total.
Question 9
1 / -0

What is common among Lithium, Sodium and Potassium?

A
These have one electron in their outermost shell

B
There are alkaline earth metals

C
These cannot form oxides

D
These are inert elements

Solution

Lithium (Li), Sodium (Na), and Potassium (K) has one electron in their outermost orbits.
They belong to group 1 and hence they are Alkali Metals and not alkali earth metal.
Lithium can form lithium oxide, Sodium forms Sodium oxide and Potassium can form Potassium peroxide in the presence of oxygen.
They are not Inert elements as they react with other compounds.
Question 10
1 / -0

\(3 \mathrm{~g}\) of activated charcoal was added to \(50 \mathrm{~mL}\) of the acetic acid solution \((0.06 \mathrm{~N})\) in a flask. After an hour, it was filtered and the strength of the filtrate was found to be \(0.042 \mathrm{~N}\). The amount of acetic acid adsorbed (per gram of charcoal) is :

A
18 mg

B
36 mg

C
42 mg

D
54 mg

Solution

Given,
\(3 \mathrm{~g}\) of activated charcoal was added to \(50 \mathrm{~mL}\) of the acetic acid solution \((0.06 \mathrm{~N})\) in a flask.
\(50 \mathrm{~mL}\) of \(0.06 \mathrm{~N}\) acetic acid solution contains \(0.06 \mathrm{~mol} / \mathrm{L}\)
\(\times \frac{50 \mathrm{~mL}}{1000 \mathrm{~mL} / \mathrm{L}} \times 60 \mathrm{~g} / \mathrm{mol}=0.180 \mathrm{~g}\) of acetic acid.
After filtration, \(50 \mathrm{~mL}\) of \(0.042 \mathrm{~N}\) acetic acid solution contains \(0.042 \mathrm{~mol} / \mathrm{L} \times \frac{50 \mathrm{~mL}}{1000 \mathrm{~mL} / \mathrm{L}}=0.126 \mathrm{~g}\) of acetic acid.
Thus, the amount of acetic acid adsorbed on \(3 \mathrm{~g}\) of activated charcoal is \(0.180 \mathrm{~g}-0.126 \mathrm{~g}=0.054 \mathrm{~g}\). This is equal to \(54 \mathrm{mg}\).
So, the amount of acetic acid adsorbed on \(1 \mathrm{~g}\) of activated charcoal is \(\frac{54 \mathrm{mg} \times 1 \mathrm{~g}}{3 \mathrm{~g}}=18 \mathrm{mg}\).