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Physical changes alter the appearance or state of a substance without changing its chemical composition, while chemical changes result in the formation of new substances with different properties.

1 mole of gas at s.t.p = 22.4 2.5 moles → 2.5 * 22.4 = 56.00 dm3

The correct answer is A. Nearly equal masses with a release of energy and radiation. However, nuclear fusion actually involves the combining of two light nuclei to form a heavier nucleus, not the splitting of a heavy element. This process releases a significant amount of energy and, in some cases, radiation. Examples of nuclear fusion include: - The sun's core, where hydrogen nuclei (protons) fuse to form helium, releasing vast amounts of energy in the process. - Fusion reactions in tokamaks and other experimental devices, where isotopes of hydrogen (such as deuterium and tritium) are fused to form helium and release energy. The correct description of nuclear fission, on the other hand, is: - The splitting of a heavy element (such as uranium or plutonium) into two or more lighter nuclei, releasing energy and radiation in the process.

Given: m = 25g n = 0.4 M = ? Using Number of moles, n = \(\frac{mass}{Molar mass}\) M = \(\frac{m}{n}\) M = \(\frac{25}{0.4}\) M = 62.5g

The element with the electron configuration 1S\(^2\)2S\(^2\)2P\(^3\) is Nitrogen.

The correct answer is C. A decrease in electropositivity. As you move across a period in the periodic table: - Atomic size decreases due to the increase in effective nuclear charge. - Nuclear charge increases due to the addition of protons. This results in: - A stronger attraction between the nucleus and electrons, making it more difficult for the atom to lose electrons. - A decrease in electropositivity (the tendency of an atom to lose electrons and form a positive ion). The other options are incorrect: A. The tendency to lose electrons actually decreases across a period. B. Chemical reactivity can increase or decrease across a period, depending on the specific elements and their electron configurations. D. The decrease in atomic size and increase in nuclear charge can actually lead to an increase in the covalent character of bonds, as the atom becomes more electronegative.

To separate sand, NH₄Cl, and NaCl, you can use sublimation to isolate NH₄Cl, followed by dissolving the remaining mixture in water, filtering out the sand, and then evaporating the solution to recover NaCl. Sublimation of NH₄Cl Heat the mixture of sand, NH₄Cl, and NaCl in a china dish covered with a funnel (plugged with cotton). NH₄Cl will sublime (transform directly from solid to gas) and condense on the cooler surface of the funnel, leaving behind a mixture of sand and NaCl. Collect the NH₄Cl from the funnel walls. Dissolving and Filtering: Add water to the remaining mixture of sand and NaCl. NaCl will dissolve in water, while sand will remain undissolved. Filter the mixture to separate the sand from the NaCl solution. 3. Evaporating the NaCl Solution: Evaporate the water from the NaCl solution, leaving behind solid NaCl crystals.

Ernest Rutherford proposed the idea of the planetary model of the atom, where electrons orbit a central nucleus, in 1911. The idea of a "planetary electron," where electrons orbit the nucleus like planets around the sun, is attributed to Ernest Rutherford; his model is often called the "planetary model" of the atom. While Bohr later refined this model by proposing fixed energy levels for electrons, the initial concept of electrons orbiting a central nucleus like planets was put forward by Rutherford. The idea of a "planetary" model of the atom, where electrons orbit the nucleus like planets around the sun, was later proposed by Niels Bohr in 1913.

Given: V\(_1\) = 30cm\(^3\) T\(_1\) = 50\(^0\)C + 273 = 323K V\(_2\) = ? T\(_2\) = 80\(^0\)C + 273 = 353K According to Charles' law: \(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) V\(_2\) = \(\frac{{353}\times{30}}{323}\) V\(_2\) = 32.79 cm\(^3\) Increase in volume = V\(_2\) - V\(_1\) = 32.79 - 30 = 2.79 cm\(^3\) Fractional increase in volume = \(\frac{{V_2} - {V_1}}{V_1}\) = \(\frac{{32.79} - {30}}{30}\) = 0.09287 = 0.093

2C\(_2\)H\(_2\)(g) + 5O\(_2\)(g) → 4CO\(_2\)(g) + 2H\(_2\)O(g) 2 moles : 4 moles n = \(\frac{mass}{molar mass}\) = \(\frac{78}{26}\) If 3 moles : 6 moles n = \(\frac{m}{M}\) m = n x M m = 6 x 44 m = 264g

All the other options given, in their atom state can be good oxidizing agents but the ions are reducing agents.

In a Leclanche cell, the anode and the cathode are a zinc rod and a carbon rod respectively. The electrolyte is an ammonium chloride solution or paste (wet or dry cell).

H\(_2\)S = 0 Bearing in mind that the oxidation state of H = + 1 H\(_2\) + S = 0 2(+1) + S = 0 + 2 + S = 0 S = 0 - 2 S = - 2

\(\frac{C_A V_A}{C_BV_B} = \frac{a}{b}\) \(\frac{0.05 \times 10}{C_B \times 25} = \frac{2}{1}\) 50C\(_B\) = 0.5 C\(_B\) = \(\frac{0.5}{50} = \frac{1}{100}\) = 0.01

Heat of combustion (\DeltaH^\circc) is the energy released when a substance is burnt in oxygen. The heat of formation (\DeltaH^\circf) is the energy released or absorbed when a substance is formed from its constituent elements. The chemical equation is H\(_2\) + 1/2 O\(_2\) → H\(_2\)O Since the heat of formation of H\(_2\) is zero (\DeltaHf(H\(_2\) ) = 0 kJ/mol), and the heat of combustion of H\(_2\) is -285.8 kJ/mol, we can use the following relationship: \DeltaHc(H\(_2\) ) = \DeltaHf( H\(_2\)O) - \DeltaHf(H\(_2\) ) Rearrange to solve for \DeltaHf( H\(_2\)O): \DeltaHf( H\(_2\)O) = \DeltaHc(H\(_2\) ) + \DeltaHf(H\(_2\) ) = -285.8 kJ/mol + 0 kJ/mol = -285.8 kJ/mol