Chemistry Test 26

Result

Chemistry Test 26

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
4 / -1

Which type of linkage is present in starch ?

A
C₁ – C₄ glucosidic linkage

B
C₁ – C₆ glucosidic linkage

C
both (a) and (b)

D
amide linkage

Solution

C₁ – C₄ glucosidic linkage is present in linear chain whereas C₁- C₆ glucosidic linkage is present in branching.
Question 2
4 / -1

Select the correct statement from the following?

A
Covalent hydrides are formed by s-block elements.

B
The stability of covalent hydrides decreases in a period from left to right

C
Co-valent hydrides have low melting and boiling points

D
Co-valent compounds are good conductor of electricity.

Solution

Covalent hydrides are formed by p-block elements with general formula MH_{18 – a} where ‘a’ is the group to which M belongs and the stability of covalent hybrids increases in a period from left to right with increase in electronegativity.
Question 3
4 / -1

The arsenious sulphide sol has negative charge. The maximum coagulating power for precipitating is of:

A
0.1N Zn(NO₃)

B
0.1 N Na₃PO₄

C
0.1 N ZnSO₄

D
0.1 N AlCl₃

Solution

More is the change more is the coagulating power.
Question 4
4 / -1

Which of the following is an alkaloid

A
Nicotine

B
Piperine

C
Coniine

D
All of these

Solution

All these are alkaloids
Question 5
4 / -1

The correct order of acidic strength is

A
Cl₂O₇> SO₃> P₄O₁₀

B
CO₂> N₂O₄ > SO₃

C
Na₂O> MgO> Al₂O₃

D
K₂O> CaO> MgO

Solution

Acidic character of oxides increases as we move from left to right in a periodic table.
Question 6
4 / -1

The solubility product of silver benzoate is 2.5x10^-13 and the acid dissociation constant of benzoic acid is 1/3.5x10^-5. Calculate the solubility of silver benzoate in a buffer solution having pH=4.

A
10^-7

B
3x10^-6

C
5x10^-7

D
10^-6

Solution

S=√k_sp(1+[H+]/k_a]
S=√2.5x10^-13(1+10x3.5)
=√2.5x10^-13x36
=3x10^-6

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemistry Test 26

Result

Chemistry Test 26

Score

-

Rank

-

SHARING IS CARING

Question 1

C1 – C4 glucosidic linkage

C1 – C6 glucosidic linkage

both (a) and (b)

amide linkage

Solution

Question 2

Covalent hydrides are formed by s-block elements.

The stability of covalent hydrides decreases in a period from left to right

Co-valent hydrides have low melting and boiling points

Co-valent compounds are good conductor of electricity.

Solution

Question 3

0.1N Zn(NO3)

0.1 N Na3PO4

0.1 N ZnSO4

0.1 N AlCl3

Solution

Question 4

Nicotine

Piperine

Coniine

All of these

Solution

Question 5

Cl2O7 > SO3 > P4O10

CO2 > N2O4 > SO3

Na2O > MgO > Al2O3

K2O > CaO > MgO

Solution

Question 6

10-7

3x10-6

5x10-7

10-6

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

C₁ – C₄ glucosidic linkage

C₁ – C₆ glucosidic linkage

0.1N Zn(NO₃)

0.1 N Na₃PO₄

0.1 N ZnSO₄

0.1 N AlCl₃

Cl₂O₇> SO₃> P₄O₁₀

CO₂> N₂O₄ > SO₃

Na₂O> MgO> Al₂O₃

K₂O> CaO> MgO

10^-7

3x10^-6

5x10^-7

10^-6