Topics in Chemistry
Gas laws in chemistry Charles's law explained Calculation questions on Charles's law Examples of Charles's law in real life Boyle's law explained Calculation questions on Boyle's law Examples of Boyle's law in real life Summary on the kinetic molecular theory of gases Postulates of kinetic theory of gases Avogadro's number explained with worked examples Mole and Avogadro's Number explained Le Chatelier's Principle: Changes in concentration and pressure in dynamic equilibrium Chemical Equilibrium: Dynamic Equilibrium in Chemistry Static and Dynamic Equilibrium explained with their differences Chemistry Scheme of Work, SS1, First Term Chemistry Scheme of Work, SS1, Second Term Chemistry Scheme of Work, SS1, Third Term Compounds in Chemistry: Characteristics of Compounds Types of Mixture: Homogenous and Heterogeneous Mixtures What are mixtures? Characteristics of mixturesAcademic Questions in Chemistry
_____ electron(s) is a term that describes the number of electron(s) in the outermost shell of an atom.
A. Outer
B. Excess
C. Valence
D. Positive
E. Negative
F. Last
_____ is the negative electrode in electrolysis.
A. Anode
B. Anion
C. Cathode
D. Cation
E. Ion
F. Electrolyte
The above diagram shows the _____ type of bond.
A. Covalent
B. Polar covalent
C. Coordinate covalent
D. Metallic
E. Van dear walls
F. Ionic
Metals are referred to as _____ in their impure state.
A. Diluted
B. Consecrated
C. Coloured
D. Ores
E. Stained
F. Strained
Metals generally have the quality to shine, glow, sparkle, glitter, reflect light and be polished. This characteristic of metals is termed _____.
A. State
B. Ductility
C. Luster
D. Malleability
E. Hardness
F. Inflorescence
The _____ spectrometry experiment conducted on isotopic elements gave a confirmation for the existence of isotopes.
A. Mole
B. Volume
C. Weight
D. Number of moles
E. Mass
F. Amount of substance
John Dalton's first atomic theory was modified based on a discovery made by _____.
A. Sir Isaac Newton
B. Albert Einstein
C. Avogadro
D. Rutherford
E. Boyle and Charles
F. Gay Lussac
Which of the following isn't an element of the periodic table.
A. J
B. Y
C. W
D. B
E. U
F. K
Charles's law states that the volume of a given mass of gas is directly proportional to the absolute temperature of the gas (in kelvin), provided the pressure remains constant.
French physicist, Jacques Charles (1746-1823) formulated this law in 1780.
Please read the details of Charles's law alongside its graph and equations here.
A tube is filled with hydrogen gas to a volume of 10dm^{3 }at a temperature of 30^{o}C. If this tube is heated to a temperature of 100^{o}C, what will be the new volume of the tube?
Step 1: List the known quantities from the question.
Initial Volume (V_{1}) =10dm^{3}
Initial Temperature (T_{1}) = 30^{o}C
Final Volume (V_{2}) = Unknown
Final Temperature (T_{2})= 100^{o}C
Step 2: When solving questions relating to Charles's law, the temperature must be in Kelvin. From our question, the temperature was given in degree celcius (^{o}C). Therefore, we need to covert the temperature from degree celcius to kelvin. To do this, we simply add 273.15 to the celcius temperature.
T_{1} = 30 + 273.15 = 303.15K
T_{2} = 100 + 273.15 = 373.15K
Step 3: Apply the formula for Charles's law to find the final volume (V_{2}). Below is the formula:
Step 4: Make V_{2} the subject of the formula.
V_{1}/T_{1} = V_{2}/T_{2}
V_{1} x T_{2} = V_{2} x T_{1}
V_{1} x T_{2} / T_{1} = V_{2}
V_{2} = (V_{1} x T_{2}) ÷ T_{1}
Step 5: Substitute values into the formula to find the final volume (V_{2}).
V_{2} = (10 x 373.15) ÷ 303.15
V_{2} = 3731.5 ÷ 303.15
V_{2} = 12.3dm^{3}
Step 6: Think about your answer.
Please read worked examples on Boyle's law here
200L of oxygen gas was compressed at a temperature of 37^{o}C to 20L. Calculate is the temperature of oxygen at the final volume.
Step 1: List the known quantities from the question.
Initial Volume (V_{1}) = 200L
Initial Temperature (T_{1})= 37^{o}C
Final Volume (V_{2}) = 20L
Final Temperature (T_{2}) = Unknown
Step 2: Covert the temperature from degree celcius to kelvin.
T_{1} = 37 + 273.15 = 310.15K
Step 3: Apply the formula for Charles's law to find the final temperature (T_{2}).
Step 4: Make T_{2} the subject of the formula.
V_{1}/T_{1} = V_{2}/T_{2}
V_{1} x T_{2} = V_{2} x T_{1}
T_{2} = V_{2} x T_{1} / V_{1}
T_{2} = (V_{2} x T_{1}) ÷ V_{1}
Step 5: Substitute values into the formula to find the final temperature (T_{2}).
T_{2} = (20 x 310.15) ÷ 200
T_{2} = 6203 ÷ 200
T_{2} = 31.02K
Step 6: Think about your answer.
Please read the introduction to gas laws here.
What volume of gas will be changed to 250mL if its temperature changes from 5K to 10K.
Step 1: List the known quantities from the question
Initial Volume (V_{1}) = ?
Initial Temperature (T_{1}) = 5K
Final Volume (V_{2}) = 250mL
Final Temperature (T_{2}) = 10K
Step 2: Apply the formula for Charles's law to find the initial volume (V_{1}).
Please read on examples of Charles's law in real life here.
Step 3: Make V_{1} the subject of the formula.
V_{1}/T_{1} = V_{2}/T_{2}
V_{1} x T_{2} = V_{2} x T_{1}
_{V1} = V_{2} x T_{1} / T_{2}
V_{1} = (V_{2} x T_{1}) ÷ T_{2}
Step 4: Substitute values into the formula to find the initial volume (V1).
V_{1} = (250 x 5) ÷ 10
V_{1} = 1250 ÷ 10
V_{1} = 125mL
Step 5: Think about your answer.
Notice that the initial volume (V1) decreases just as the initial temperature (T_{1}) decreases. This is in accordance with Charles's law.
Please read on the kinetic theory of gases here.
A noble gas occupies 3000mL of tube at a temperature of 373.15K. What volume will it occupy in litres at 100^{o}C?
Step 1: List the known quantities from the question
Initial Volume (V_{1}) = 3000mL
Initial Temperature (T_{1}) = 373.15K
Final Volume (V_{2}) = ?
Final Temperature (T_{2}) = 100^{o}C
Step 2: Covert the final temperature (T_{2}) from degree celcius to kelvin.
T_{1} = 100 + 273.15 = 373.15K
Step 3: Since our answer (final volume) is required in litres, we should convert our initial volume from millilitres to litres by dividing it by 1000.
V_{1} = 3000 ÷ 1000 = 3L
Step 4: Apply the formula for Charles law to find the final volume (V_{2}).
Step 5: Make V_{2} the subject of the formula.
V_{1}/T_{1} = V_{2}/T_{2}
V_{1} x T_{2} = V_{2} x T_{1}
V_{1} x T_{2} / T_{1} = V_{2}
V_{2} = (V_{1} x T_{2}) ÷ T_{1}
Step 6: Substitute values into the formula to find the final volume (V_{2}).
V_{2} = (3 x 373.15) ÷ 373.15
V_{2} = 1119.45 ÷ 373.15
V_{2} = 3L
Step 6: Think about your answer.
You can read on Avogadro's number explained with worked examples here.
At -30^{o}C, a gas has a volume of 500mL. What will be its new temperature if the volume decreases by 450mL?
Step 1: List the known quantities in the question.
Initial Volume (V_{1}) = 500mL
Initial Temperature (T_{1}) = -30^{o}C
Final Volume (V_{2}) = 500mL - 450mL -> 50mL
Final Temperature (T_{2}) = Unknown
Step 2: Covert initial temperature (T_{1}) to kelvin.
T_{1} = -30 + 273.15 = 243.15K
Step 3: Make T_{2} the subject of the formula. You can do this by cross multiplying.
V_{1}/T_{1} = V_{2}/T_{2}
V_{1} x T_{2} = V_{2} x T_{1}
T_{2} = V_{2} x T_{1} / V_{2}
T_{2} = (V_{2} x T_{1}) ÷ V_{1}
Step 4: Substitute values into the formula to find the final temperature (T_{2}).
T_{2} = (50 x 243.15) ÷ 500
T_{2} = 12157.5 ÷ 500
T_{2} = 24.32K
You can read on Graham's law of diffusion with worked examples here.
Step 5: Think about your answer.
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Amazing facts in Chemistry
Plastic and Glass can decompose, but not in our lifetimes. It takes an average time of 450 years for plastics to decompose. As for the decomposition of glasses, it takes about 4,000 years
The only letters that failed to appear on the periodic table are letters:
Gold and copper are the only two non-silvery colored metals.
Copper is the only metal that is naturally antibacterial. For this reason, some children utilize 'copper water bottles' in schools
Water freezes faster when it’s warm than when it’s cold
Most element in their pure state exists physically in different forms. For instance, pure carbon can exist as both diamond and graphite. This phenomenon is called allotropy
If you pour a handful of salt into a full glass of water, the water level will go down rather than overflowing the glass.
Similarly, if you mix half liter of water and half litre of alcohol, the total volume of the liquid will be les than one litre
NOTABLE POINTS IN Chemistry
The periodic table, also called periodic table of elements or Mendeleev's table, is a table that shows an organized arrangement of the 118 chemical elements according to their atomic number.
Out of the 118 elements; elements 1 - 94 are present in nature while elements 95 - 118 are synthesized artificially.
The manner at which elements are arranged on this table reveals some similarities in their electronic configurations and chemical properties.
A compound composed of iron (Fe) and oxygen (O) was analyzed and found to contain 69.94% iron and 30.06% oxygen. Find the empirical formula of the compound. (Molar mass of Fe=55.85, O=16)
Step 1: Identify the given parameter from the question.
Fe = 69.94%, O = 30.06%.
Empirical formula = Fe_{?}O_{?}
Step 2: Convert the percentages to gram. (just attribute grams to the %).
Step 3: To get the mole ratio of each element, convert the gram to moles using the formula (mole = mass/molarmass). Please merorize this formula because we always work with moles in emperical formula.
Mole of Fe: 69.94/55.85 = 1.252mol
Mole of O: 30.06/16 = 1.879mol
Step 4: Divide both sides by the smallest mole ratio.
Iron has the smallest mole ratio in our case, therefore: 1.252/1.252 = 1, 1.879/1.252 = 1.5
We now have the formula = Fe_{1}O_{1.5}
Step 5: Multiply each of the moles by the smallest whole number that will convert each into a whole number. (In our case, the number '2' is the smallest whole number that will make '1.5' and '1' whole numbers when multiplied by it.
For iron (Fe), we will have 1 x 2 = 2
For oxygen (O), we will have 1.5 x 2 = 3
Step 6: Write the empirical formula.
The empirical formula= Fe_{2}O_{3}
Iron(III)tetraoxosulphate(VI)
In chemistry, hydrocarbons can be classified as either aliphatic or aromatic. Recently, both classifications of hydrocarbon were based on their structure rather than their origin.
Aliphatic hydrocarbons are put into three main groups according to the types of bonds they possess. These are:
Alkanes
Alkenes
Alkynes
They are shown in the image below:
It's important to note the followings:
Alkanes have single bonds (only) in their structures.
Alkenes always have a carbon-carbon double bond present in their structure.
Alkynes always have a carbon-carbon triple bond present in their structure.
Aromatic hydrocarbons are classified into:
Organic chemistry is the study of carbon and it's compounds.
Carbon is the focus of organic chemistry because it has a wide chemical diversity in the sense that it can combine with other carbon atoms to form a long chain of carbon molecule. This ability and process whereby carbon can form a long chain of itself is called catenation.
A major challenge encountered when calculating molecular mass is that it becomes difficult or impossible to calculate especially when the relative molecular mass of large molecules, polymers and macromolecules are involved.
Examples of large molecules (with indefinite molecular masses) include carbohydrates, cellulose and complex sugars.
The large molecules (above) have no specific chemical formula throughout their volume.
Understand that Relative Molecular Mass prove to be useful only when we calculate substances with small and definite molecular sizes. This was proven through the modifications of Dalton's atomic theory.
Please read on Dalton's atomic theory and its modifications here.