Interpretation:
The electrostatic potential energies between the given charges in the given statement should be calculated by using the concept of electrostatic energy.
Concept Introduction:
Energy is the ability to do work or transfer heat where work is the movement of a body using some force. The SI unit of energy is joule (
Electrostatic energy is potential energy which results from the interaction of charged particles. Oppositely charged particles attract each other and particles of like charges repel each other. The magnitude of the resulting electrostatic potential energy is proportional to the product of the two charges (
If the charges
To calculate: Determine the
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Chemistry: Atoms First
- Consider the reactions of silver metal, Ag(s), with each of the halogens: fluorine, F2(g), chlorine, Cl2(g), and bromine, Br2(l). What chapter data could you use to decide which reaction is most exothermic? Which reaction is that?arrow_forwardFor two adjacent ions, the net potential energy is: А В + rn EN - - r where A, B and n are constants, r is in nm and E is in eV. (a) Find the expression for the bonding energy Eg in terms of A, B and n. (b) For two pairs of ions, with A = 1.436, B = 5.86 x 106 and n = 9, solve for ro and Eg.arrow_forwardFor the following question , I did not get why is the correct answer 'A'arrow_forward
- Given these reactions, where X represents a generic metal or metalloid 1) H, (g) + ¿0,(g) –→ H,O(g) AH = -241.8 kJ 2) X(s) + 2 Cl,(g) > XCl,(s) ΔΗ +153.9 kJ 3) H,(g) + ¿Cl, (g) → HCI(g) AH3 = -92.3 kJ 4) X(s) + O2(g) - XO,(s) AH4 = -520.3 kJ 5) Н, О(@) — Н,о) AH5 = -44.0 kJ what is the enthalpy, AH, for this reaction? XCI, (s) + 2 H,O(1) - · XO,(s) + 4 HCI(g) -995 ΔΗ- kJ Incorrectarrow_forwardcalculate the reaction energy of the following reactions and then determine if they are endothermic or exothermic: 2F2(g) + 2H2O => 4HF + O2arrow_forwardWrite the reaction for the enthalpy of formation of potassium perchlorate.arrow_forward
- Using the table of standard formation enthalpies that you'll find under the ALEKS Data tab, calculate the reaction enthalpy of this reaction under standard conditions: 2C,H(9)+70,(9)→4CO,(g)+6 H,0(1)arrow_forwardusing the delta H reaction equation and standard enthalpies of formation, calculate the enthalpy change of the decomposition of hydrogen peroxide into water that occurs in the presence of lightarrow_forwardPhysical constants that you may need: -19 Elementary Charge e = 1.60218 × 10¯ C Vacuum Permittivity 60 = 8.85419 × 10-12 Jm Cesium is the most reactive metal in the periodic table. It reacts vigorously with molecular compounds such as water or chlorine gas to produce ionic compounds. The dissociation energy of gaseous cesium chloride into neutral atoms is ▲Ed 439 kJ/mol. = CsCl (g) Cs (g) + Cl (g) AE = AEd Potential Energy Interatomic distance r AEd = 439 kJ mol Estimate the equilibrium distance re in gaseous cesium chloride. Report the value in picometers. lonization energy of cesium: 375.5 kJ mol Electron affinity of chlorine: 348.7 kJ mol 1220 pm xarrow_forward
- Given the following ræactions and their enthalpies: ΔΗ (k/mol), +436 2H(g) H2(g) 0,(g) - 20(g) +495 H(g) + 02(g) Н.О(g) -242 (a) Devise a way to calculate AH for the reaction H,O(g) → 2H(g) + O(g) (b) From this, estimate the H-O bond energy.arrow_forwardHess's Law is a useful way for chemists to predict the enthalpy change for a reaction. Use your understanding of energy changes to explain how Hess's Law works. Use the following example as part of your explanation (but you don't have to solve it!). 8. How can Hess's law be applied to determine the enthalpy change for the decomposition of potassium bicarbonate: 2 KHCO, (9) – K,CO; (s) + H,O () + CO2 (2) AH = -963 kJ (1) K (9) + ½ H, (e) Ce +0, (9)→ KHCO; (5) 2 (g) AH = -1150 kJ (2) 2K (9) + C + 02 (e) → K,C03 (6) AH = -242 kJ (3) H2() + ½ O2() H,O (g) AH = -394 kJ (4) C (6)+02 (g) - CO2 (2)arrow_forwardWhich one of these processes corresponds to the lattice energy of NaCl(s)? O NaCI(s) → Na*(g) + CI (g) O Na(s) + ½ Cl2(g) NaCl(s) O Na*(g) + CI(g) → NaCl(s) O NaCl(s) → Na(s) + ½ Cl2(g)arrow_forward
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