Scientific notations/Exponents forms test

Convert the following into scientific notations/exponent form.

1.                  0.001

2.                  2,010

3.                  22.50

4.                  0.00022

5.                  72.0

6.                  233.0

7.                  0.0251

8.                  0.52

9.                  98,000

10.              0.0025

Answers:

1.                  1.0 x 10^-3

2.                  2.01 x 10³

3.                  2.25 x 10¹

4.                  2.2 x 10^-4

5.                  7.2 x 10¹

6.                  2.33 x 10²

7.                  2.51 x 10^-2

8.                  5.2 x 10^-1

9.                  9.8 x 10⁴

10.              2.5 x 10^-3

Organic chemistry IUPAC Nomenclature Test class XI

Basic concept of chemistry (class XI) Chemistry

Class: XI                                                                   Subject: Chemistry

Total Marks: 15                                                       Chapter:1

Choose correct options:

1.         9 grams of water decompose to give                                                   (1)

 1g hydrogen and 8g oxygen

 8g oxygen and 1g hydrogen

 4g oxygen and 16g hydrogen

 2g hydrogen and 1g oxygen

2.         100 g of CaCO₃ on heating gave 44 g of CO₂ and 56 g of CaO. The observation is in agreement with the:                                                                  (1)

 Law of conservation of mass

 Law of constant proportion

 Law of multiple proportion

 Law of reciprocal proportion

3.         In carbon dioxide ratio of mass of carbon to mass of oxygen is always 3:8 . This is in accordance with:                                                                   (1)

 Law of conservation of mass

 Law of constant proportion

 Law of multiple proportion

 Law of reciprocal proportion,

4.         The atomic masses and chemical properties of Sulphur and phosphorus are different this can be explained on the basis of                              (1)

 Law of constant proportions

 Law of conservation of mass

 Dalton's Atomic theory

 Laws of Chemical combination

5.         Hydrogen and oxygen combine in the ratio 1:8 by mass to form water. How much mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?                                                                                        (1)

 8 g

 16 g

 24 g

 32 g

6.         Dalton's atomic theory provided explanation for                                 (1)

 Law of conservation of mass

 Law of definite proportions

 Both law of conservation of mass and law of definite proportions

 Atomic numbers

7.         Which postulate of Dalton's Atomic Theory can explain the law of Definite Proportion                                                                                              (1)

 All matter is made of very tiny particles called atoms

 Atoms of a given element are identical in mass and chemical properties.

 Atoms combine in ratio of small whole numbers to form compounds

 The relative number and kinds of atoms are constant in a given compound.

8.         What is percentage composition of oxygen in NO^3-                                                          (1)

9.         Calculate the molar mass of Ca (NO₃)₂.                                                (1)

10.        How many moles are in 26.9g of C₁₁H₁₂O₁₁.                                          (1)

11.        Find the percentage composition of KMnO₄.                                       (2)

12.        A compound has a percentage composition of 46.681% Nitrogen and 53.32 % Oxygen. It’s molecular mass is 59.98 amu. What is the empirical formula? What is it’s molecular formula?                                                                                       (3)

Molecular Mass Calculations (Chemistry)

Class: XI                                                                      Subject: Chemistry

1.                  Calculate the molecular mass of the following compounds

(a)    CuSO₄.5H₂O

(b)   K₂Cr₂O₇

(c)    Pb₃(PO₄)₂

(d)   NaNH₄HPO₄.4H₂O

(e)    FeSO₄(NH₄)2SO₄.6H₂O

ELECTRONIC CONFIGURATION SCHEME