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

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

Take a straight line(both vertical and horizontal) to touch the straight line. Acceleration = \(\frac{\text{change in velocity}}{\text{time}}\) Acceleration = \(\frac{8.7 - 2}{10 - 0}\) = \(\frac{6.7}{10}\) = 0.67m/s\(^2\).

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

Given: Force \( P = 20 \, \text{N} \), Distance \( MN = 0.4 \, \text{m} \) Moment About Point N: \(\text{Moment}_N = P \times MN = 20 \, \text{N} \times 0.4 \, \text{m} = 8 \, \text{Nm}\) Moment About Point O: \(\text{Distance from O} = 0.5 \times \cos(36.87^\circ) \approx 0.4 \, \text{m}\) \(\text{Moment}_O = P \times \text{Distance from O} = 20 \, \text{N} \times 0.4 \, \text{m} = 8 \, \text{Nm}\) Difference Between Moments: \(\text{Difference} = \text{Moment}_N - \text{Moment}_O = 8 \, \text{Nm} - 8 \, \text{Nm} = 0 \, \text{Nm}\)

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

To remove ammonium chloride from the mixture of sand, sugar, and 2 M ammonium chloride solution, the following steps are taken: Filtration: This is the first step used to separate the sand (an insoluble solid) from the liquid phase. The sugar and ammonium chloride remain dissolved in the solution and pass through the filter as the filtrate. Sublimation: After filtering, the solution containing water, sugar, and ammonium chloride is processed. Heating the mixture allows the water to evaporate and the ammonium chloride to sublime (transition directly from a solid to a gas). This process effectively isolates the ammonium chloride from the sugar, which does not sublime.

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

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

To determine the formula of the oxide, let's analyze the given information: Stoichiometrically, 2 X + 3/2 O\(_2\) → Oxide Given : 1. 1 litre of monoatomic gas X combines with 1.5 litres of an oxide. 2. The reaction is a combination reaction, where X reacts with oxygen to form an oxide. If the atom of the elements are X and O\(_2\) Reacted volumes were given as 1dm\(^3\) 1.5dm\(^3\) Stoichiometrically, moles of molecules 2 moles 3 moles X(monoatomic) and oxygen(diatomic) 2 x 1 3 x 2 = 6 (3 moles of O\(_2\) = 6 oxygen atoms) Divide ratios by 2 \(\frac{2}{2}\) \(\frac{6}{2}\) The formula ratio : 1 : 3 Therefore, the formula of the oxide is XO\(_3\)

Given: C\(_1\) = 2.2 M , V\(_1\) = 1 Litre , C\(_2\) = ? V\(_2\) = 1 + 10 Litres ( because the volume of water has increased) = 11 Litres Using the relationship C\(_1\)V\(_1\) = C\(_2\)V\(_2\) 2.2 X 1 = C\(_2\) X 11 C\(_2\) = \(\frac{2.2}{11}\) = 0.20 M - Option E

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

Van der Waals noticed the behavior of real gases at low temperature and high pressure possesses weak but short range forces.

Mg + 2HCl → MgCl2 + H2 6 molar contains 6 Χ 36.5 gms 1000 cc Solution i.e 6 Χ 36.5/2gms in 500 cc = 3 Χ 36.5 24 gms of mg will react with 2 Χ 36.5 HCl to liberate 2 gm of hydrogen ∴ 6 gms will liberate 6 Χ 2/24 Χ 1 = 0.5 gm of H2

The kinetic theory of gases assumes that gas particles are negligible in size: Small size: The total volume of the gas particles is negligible compared to the volume of the container. Random motion: The particles are in constant, random motion. No intermolecular forces: There is no force of attraction between the particles at normal temperatures and pressures. Elastic collisions: Collisions between particles are elastic, meaning there is no net loss of energy. Kinetic energy: The kinetic energy of the particles is proportional to temperature.