The first ionization energy for rubidium is +403.0 kJ mol. How much energy (as kJ) would be required to convert 19.10 g of gaseous rubidium to its gaseous +1 monatomic ion at constant temperature? Do not write units, only enter numerical values. Use two digits at most for decimal numbers, and use dot (.) as decimal separator (i.e. two-point- fifty should be entered as 2.50). Do not use comma (,) as decimal separator.

The first ionization energy for rubidium is +403.0 kJ mol. How much energy (as kJ) would be required to convert 19.10 g of gaseous rubidium to its gaseous +1 monatomic ion at constant temperature? Do not write units, only enter numerical values. Use two digits at most for decimal numbers, and use dot (.) as decimal separator (i.e. two-point- fifty should be entered as 2.50). Do not use comma (,) as decimal separator.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

help please answer in text form with proper workings and explanation for each and every part and steps with concept and introduction no AI no copy paste remember answer must be in proper format with all working
Why does a calcium atom become more stable when it loses two electron? For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac).
Q2: a- Explain how such a model uses potentials such as a square well and a harmonic oscillator to try and predict the magic numbers. b- Describe what causes the existence of "magic numbers" for protons and neutrons in nuclei. Why are nuclei which have these magie numbers more stable than other nuclei? Draw any analogies with the energy levels of atomic electrons. c- Discuss briefly the failures of the shell model. of
Q1 and Q2 please
Calculate the energy of 1.0 amu in joules using the Einstein s relationship. (1 J= 1 kg.m /s , 1g =6.02x1023 amu) E = C2 (deltam)
Please I want answer for this question by typing. Many Thanks Atomic size is one of the most important periodic property of elements in the periodic table. Prepare a report on the effect of atomic size on ionization energy, electron affinity, electronegativity and metallic character of elements. (Word Limit 200)
(3a-104-17) Would you agree to buy 1.23 x 1021 platinum atoms for $17? Explain why or why not, showing all work. The current price of platinum is 30.71 dollars/gram.
Conside following data for lithium: atomic mass 6.941 electronegativity electron affinity 59.6 ionization energy heat of fusion 520.2 0.98 3.0 (1) Li(g) + e Li (g) + (2) Li (g) + e g mol - You may find additional useful data in the ALEKS Data tab. kJ mol Does the following reaction absorb or release energy? kJ mol Li (g) kJ mol Is it possible to calculate the amount of energy absorbed or released by reaction (1) using only the data above? If you answered yes to the previous question, enter the amount of energy absorbed or released by reaction (1): Does the following reaction absorb or release energy? Is it possible to calculate the amount of energy absorbed or released by reaction (2) using only the data above? If you answered yes to the previous question, enter the amount of energy absorbed or released by reaction (2): OOO release absorb Can't be decided with the data given. yes no kJ/mol release absorb Can't be decided with the data given. yes no ☐kJ/mol
The energy of an electron in a one-electron atom or ionequals 1-2.18 * 10-18 J2 aZ2n2 b, where Z is the nuclear chargeand n is the principal quantum number of the electron. Estimatethe first ionization energy of Li.
E MasteringChemistry: Post Lecture x P Course Home A session.masteringchemistry.com/myct/itemView?assignme 0 Turnitin A ALEKS -- Assessme. f (49) Facebook GG A Games on Sale.
i need the answer quickly
Given the following processes and their respective enthalpies in kJ/mol: Li(s) --> Li(g); DHs = +161 Li(g) --> Li+(g) + e–; IE1 = +520 F2(g) --> 2F(g); BE = +160 F(g) + e– --> F-(g); EA = -328 Li+(g) + F-(g) --> LiF(s); UL = x If the reaction: Li(s) + ½F2(g) --> LiF(s) has DHf= -616.0 kJ/mol, deduce the value of x for UL (lattice energy in kJ/mol) in LiF. (A) -1129 (B) -1049 (C) -183 (D) -103