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The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is [2003]a)9.92 × 105 Pab)9.92 × 108 Pac)9.92 × 107 Pad)9.92 × 106 PaCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is [2003]a)9.92 × 105 Pab)9.92 × 108 Pac)9.92 × 107 Pad)9.92 × 106 PaCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is [2003]a)9.92 × 105 Pab)9.92 × 108 Pac)9.92 × 107 Pad)9.92 × 106 PaCorrect answer is option 'B'. Can you explain this answer?.

Solutions for The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is [2003]a)9.92 × 105 Pab)9.92 × 108 Pac)9.92 × 107 Pad)9.92 × 106 PaCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11. Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.

Here you can find the meaning of The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is [2003]a)9.92 × 105 Pab)9.92 × 108 Pac)9.92 × 107 Pad)9.92 × 106 PaCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is [2003]a)9.92 × 105 Pab)9.92 × 108 Pac)9.92 × 107 Pad)9.92 × 106 PaCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is [2003]a)9.92 × 105 Pab)9.92 × 108 Pac)9.92 × 107 Pad)9.92 × 106 PaCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is [2003]a)9.92 × 105 Pab)9.92 × 108 Pac)9.92 × 107 Pad)9.92 × 106 PaCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm–3, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is [2003]a)9.92 × 105 Pab)9.92 × 108 Pac)9.92 × 107 Pad)9.92 × 106 PaCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 11 tests.