Learn more about the properties, composition, and structure of substances (elements and compounds) with these Chemistry questions and answers. This Test can be used by students preparing for Chemistry in JAMB, WAEC, NECO or Post UTME.
25%
12%
75%
50%
Correct answer is C
The molar mass of methane (CH4) is calculated by adding the atomic masses of carbon (approximately 12 g/mol) and hydrogen (approximately 1 g/mol). Therefore, the molar mass of CH4 is 12 g/mol + 4 x 1 g/mol = 16 g/mol. The mass percentage of carbon in CH4 is given by (mass of carbon / total mass) x 100% = (12 g / 16 g) x 100% = 75%.
What type of reaction is involved in the formation of alkanols from alkenes?
Elimination reaction
Redox reaction
Substitution reaction
Addition reaction
Correct answer is D
Alkanols can be formed from alkenes through an addition reaction. In this reaction, the carbon-carbon double bond of the alkene is broken, and a hydroxyl group (-OH) is added to one of the carbon atoms, resulting in the formation of an alcohol.
What is the common name for ethanoic acid?
Acetic acid
Butyric acid
Propionic acid
Formic acid
Correct answer is A
Ethanoic acid is commonly known as acetic acid. It is the main component of vinegar and is used in various industrial processes.
Chromatography
Filtration
Decantation
Distillation
Correct answer is A
Chromatography is a separation technique used to separate and analyze components of a mixture based on their differential partitioning between a stationary phase (e.g., paper or a column) and a mobile phase (e.g., solvent). Different components move at different rates, leading to separation.
C4H8O4
CH2O
C2H4O2
C3H6O3
Correct answer is B
To determine the empirical formula, we first need to convert the percentages to moles by assuming a convenient mass for the sample. Let's assume we have 100 grams of the compound.
Mass of carbon = 40.00 grams
Mass of hydrogen = 6.67 grams
Mass of oxygen = 53.33 grams
Now, we calculate the moles of each element:
Moles of carbon = 40.00 g / molar mass of carbon = 40.00 g / 12.01 g/mol ≈ 3.33 mol
Moles of hydrogen = 6.67 g / molar mass of hydrogen = 6.67 g / 1.01 g/mol ≈ 6.60 mol
Moles of oxygen = 53.33 g / molar mass of oxygen = 53.33 g / 16.00 g/mol ≈ 3.33 mol
Now, we find the smallest whole-number ratio of the moles:
C ≈ 3.33 / 3.33 ≈ 1
H ≈ 6.60 / 3.33 ≈ 2
O ≈ 3.33 / 3.33 ≈ 1
Thus, the empirical formula is CH2O.