Question Library

The IUPAC name of the compound CF\(_3\)CHBrCl is 2-bromo-2-chloro-1,1,1-trifluoroethane. This compound is famously known by its non-proprietary name, Halothane, and was widely used as an inhalation anesthetic. Breakdown of the Name To arrive at the correct IUPAC name, we follow standard nomenclature rules: Identify the Parent Chain: The longest carbon chain contains two carbons, making the parent alkane ethane. Numbering the Chain: We number the carbons to give the substituents the lowest possible locants. If we start from the carbon with the three fluorine atoms, the locants are (1,1,1,2,2). If we start from the carbon with bromine and chlorine, the locants are (1,1,2,2,2). In this case, both directions yield the same set of locants. However, IUPAC rules state that if numbering is equivalent from both ends, we give the lower number to the substituent that comes first alphabetically. Since Bromo and Chloro come before Fluoro, the carbon attached to Br and Cl is designated as Carbon-2 (wait, let me re-check that logic). Actually, per the lowest locant set rule, we prioritize the side with more substituents to keep the numbers lower. The carbon with three fluorines gets "1" so the set is (1,1,1,2,2) rather than (1,2,2,2,2). Correction: In the specific case of Halothane, the standard IUPAC assignment labels the carbon with the three fluorines as C1, resulting in 2-bromo-2-chloro-1,1,1-trifluoroethane

No expanded working has been attached for this question yet. The verified answer is Option D.

The chemical reaction between magnesium ribbon and oxygen is a combustion reaction that produces magnesium oxide (MgO). Burning Magnesium ribbon in air removes oxygen, because as Magnesium burns in air, it reacts with oxygen in the air to form Magnesium oxide. The chemical equation for this reaction is 2Mg(s)+O\(_2\)(g) → 2MgO(s)

The best starting material for the preparation of oxygen through heating trioxonitrate(V) (nitrates) would involve a metal that facilitates the decomposition of the nitrate to produce oxygen, coincidentally, Pb (Lead) falls into that category. Lead(II) nitrate (Pb(NO\(_3\))\(_2\)) decomposes upon heating to produce oxygen, nitrogen dioxide, and lead(II) oxide. This reaction is often used in laboratory settings to generate oxygen. 2Pb(NO\(_3\))\(_2\) → 2PbO + 4NO\(_2\) + O\(_2\) This makes lead (Pb) a suitable choice for generating oxygen from nitrates.

No expanded working has been attached for this question yet. The verified answer is Option C.

The Group 2 metals become more reactive towards water as you go down the Group.

When Chlorine water is exposed to bright sunlight, the products formed are hydrochloric acid (HCl) and oxygen gas O\(_2\).

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option A.

Sulphur trioxide (SO\(_3\)) is not directly dissolved in water to produce sulphuric acid (H\(_2\)SO\(_4\)) because the reaction is highly exothermic and produces a fine mist of sulphuric acid, which is difficult to control and condense. Instead, SO\(_3\) is absorbed in concentrated sulphuric acid (oleum) to form oleum, which is then diluted with water to produce the desired sulphuric acid concentration.

Rutherford's initial atomic model, though crucial for understanding the nucleus, didn't explicitly include neutrons. Rutherford's model, developed in 1911, identified a small, positively charged nucleus at the atom's center, but it only included positively charged protons within this nucleus. The neutral neutron was discovered later, by James Chadwick, in 1932. Rutherford's experiment involved bombarding a thin layer of gold with alpha particles (helium nuclei). The results showed that most alpha particles passed through the gold foil with little deflection and a small fraction of alpha particles were deflected at large angles. Rutherford's observations led him to propose the nuclear model of the atom, which states that: 1. Atoms have a small, dense, positively charged nucleus (containing most of the atom's mass). 2. The nucleus is surrounded by a cloud of electrons. The deflection of alpha particles in Rutherford's experiment was indeed caused by the massive, positively charged nucleus. The nucleus's positive charge and concentrated mass led to the scattering of alpha particles at large angles. The nucleus is not described as "very large" in Rutherford's theory. Instead, it's the empty space surrounding the nucleus that's large, allowing most alpha particles to pass through with minimal deflection. These also make option D incorrect, leaving us with option A and D as the answers, although more likely option D.

For better understanding, try to identify what elements X and Y are: Elements X with electronic configurations: 1s\(^2\)2s\(^2\)2p\(^4\) is Oxygen (Adding the electrons together as 2+2+4= 8) Elements Y with electronic configurations 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\)3p\(^1\) is Aluminium ( Adding the electrons together as 2+2+6+2+1 = 13) When Aluminium being a metal reacts with Oxygen being a non-metal, there is a transfer of electrons from metals to non-metals as 3 oxygen atoms reacts with 2 Aluminium atoms thereby forming Al\(_2\)O\(_3\) ⇒ Y\(_2\)X\(_3\).

Parameters given: V\(_1\) = 200 cm\(^3\); T\(_1\) = 27 + 273 = 300K. V\(_2\) = 150 cm\(^3\); T\(_2\) = ? P\(_2\) = 2P\(_1\). Using the Ideal Gas Equation: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) \(\frac{P_1 \times 200}{300} = \frac{2P_1 \times 150}{T_2}\) \(\frac{2}{3} = \frac{300}{T_2}\) T\(_2\) = 450K. = 450K - 273K = 177^\circC 177 - 27 = 150 That is the amount by which T1 must be increased.

2CO + O\(_2\) → 2CO\(_2\) From the equation, we see that 2 volumes of CO react with 1 volume of O\(_2\) to produce 2 volumes of CO\(_2\). Given: Volume of CO = 50 cm³ Volume of O\(_2\) = 100 cm³ First, let's find the limiting reactant: CO: 50 cm³ / 2 = 25 cm³ of O\(_2\) required (based on the stoichiometry) However, we were told that 50cm\(^3\) of carbon(II)oxide was exploded in 100cm\(^3\) of oxygen, which means Oxygen is in excess (100 cm³ ), then CO is the limiting reactant, it will be completely consumed. Now, let's calculate the volume of CO\(_2\)` produced: 50 cm³ of CO produces 50 cm³ of CO\(_2\) i.e (1:1 ratio) The resulting mixture will contain 50 cm³ of CO\(_2\) and the remaining O\(_2\) (100 cm³ - 25 cm³ ) = 75 cm³ Total volume of the resulting mixture = Volume of product (CO\(_2\)) + Volume of the remaining O\(_2\) = 50 cm\(^3\) + 75 cm\(^3\) = 125 cm\(^3\) So, the volume of the resulting mixture at the end of the reaction is 125 cm³.

When ammonia and hydrogen ion go into bonding, they form ammonium ion with a one dative covalent bond and three covalent bonds.