This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Luma, let's knock this out. Question 25: Step 1: Analyze the given ionization energies. The ionization energies are: IE1 = 1290 kJ/mol IE2 = 3370 kJ/mol IE3 = 5320 kJ/mol IE4 = 7440 kJ/mol IE5 = 10900 kJ/mol IE6 = 3010 kJ/mol IE7 = 70900 kJ/mol Successive ionization energies should always increase because it takes more energy to remove an electron from an increasingly positive ion. The value for IE6 (3010 kJ/mol) is lower than IE5 (10900 kJ/mol), which indicates a likely typo in the question. However, we can still identify the significant jump that indicates the removal of a core electron. Step 2: Identify the largest jump in ionization energy. A very large increase in ionization energy occurs when an electron is removed from a stable, filled inner electron shell (core electron) compared to removing a valence electron. Let's look at the ratios of successive ionization energies, focusing on the largest increase: Ratio IE2/IE1 = 3370/1290 ≈ 2.61 Ratio IE3/IE2 = 5320/3370 ≈ 1.58 Ratio IE4/IE3 = 7440/5320 ≈ 1.40 Ratio IE5/IE4 = 10900/7440 ≈ 1.46 Ratio IE7/IE6 = 70900/3010 ≈ 23.55 (This is the largest relative jump, despite the problematic IE6 value). Step 3: Determine the number of valence electrons. The extremely large jump between IE6 and IE7 indicates that the first 6 electrons are valence electrons, and the 7th electron is a core electron. Therefore, the atom has 6 valence electrons. Step 4: Determine the electronic configuration. The element is between Li (atomic number 3) and Ne (atomic number 10). An element in the second period with 6 valence electrons is Oxygen (O), which has an atomic number of 8. The electronic configuration for Oxygen is 1s^2 2s^2 2p^4. The question asks for the electronic configuration of the atom, typically referring to the valence shell configuration for elements in the second period. The valence electronic configuration is 2s^2 2p^4. Step 5: Match with the given options. The configuration 2s^2 2p^4 matches option D. The final answer is D. 2s^22p^4. Question 26: Step 1: Determine the number of half-lives. The fraction of the radioisotope remaining is (1)/(16) of its original value. The formula for radioactive decay is: (N)/(N_0) = ((1)/(2))^n where N is the amount remaining, N_0 is the initial amount, and n is the number of half-lives. Substitute the given fraction: (1)/(16) = ((1)/(2))^n We know that 2^4 = 16, so ((1)/(2))^4 = (1)/(16). Therefore, the number of half-lives, n, is 4. Step 2: Calculate the half-life. The total time elapsed is given as 96.8 days. The total time is also equal to the number of half-lives multiplied by the half-life period (t_1/2): t = n × t_1/2 Substitute the known values: 96.8 days = 4 × t_1/2 Solve for t_1/2: t_1/2 = 96.8 days4 t_1/2 = 24.2 days Step 3: Match with the given options. The calculated half-life of 24.2 days matches option B. The final answer is B. 24.2 days. Send me the next one 📸