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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.

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

Given: The equation below A\(_{(s)}\) + 2B\(_{(g)}\) → 2C\(_{(aq)}\) + D\(_{(g)}\) According to Le Chatelier's Principle: Pressure changes affects only the gaseous reactants and products There must be different amount of gaseous particles on both sides. Generally; a decrease in the pressure shifts the equilibrium position in favor of the side with larger volume of gases. I.e rates of backward reaction. NB: This is a correction to the video explanation.

Given: xP + yQ ⇌ mR + nS Let us assume they are all in aqueous solution k = \(\frac{[R]^m[S]^n}{[P]^x[Q]^y}\)

Option B is the only oxidation process because there was an increase in oxidation state of Oxygen from - 2 to 0.Also, there was loss of electrons. The other options are reduction reactions as they involve gain of electrons as well as decrease in oxidation number.

Potassium dichromate is an oxidising agent in organic chemistry. It is used to oxidize alcohols. In an aqueous solution the color change exhibited can be used to test for distinguishing aldehydes from ketones. Aldehydes reduce dichromate from the +6 to the +3 oxidation state, changing color from orange to green. This color change arises because the aldehyde can be oxidized to the corresponding carboxylic acid.