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AS Chemistry : Avogradro's Constant

Hello TSR,

Recently I encountered the following problem :

http://prntscr.com/lzt5ia

Thank you in advance, your cooperation will be highly appreciated.

Warm regards,
OmarEdExcel.
Original post by OmarEdExcel
Hello TSR,

Recently I encountered the following problem :

http://prntscr.com/lzt5ia

Thank you in advance, your cooperation will be highly appreciated.

Warm regards,
OmarEdExcel.

Number of moles of CO2 = 3.06/30.6 = 0.1 mol

Number of moles of atoms = 0.3 mol

Hence number of atoms = 0.3 x 6 x 10^23 hence C
Reply 2
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OmarEdExcel
OP
9
Original post by BobbJo
Number of moles of CO2 = 3.06/30.6 = 0.1 mol

Number of moles of atoms = 0.3 mol

Hence number of atoms = 0.3 x 6 x 10^23 hence C


Oh yeah, I've been acknowledged about this before, this really rings a bell. By the way, there are also some common questions that come up telling us to count the number of ions present in one compound, say Ca(OH)2. How would that be calculated?
(edited 5 years ago)
Original post by OmarEdExcel
Oh yeah, I've been acknowledged about this before, this really rings a bell. By the way, there are also some common questions that come up telling us to count the number of ions present in one compound, say Ca(OH)2. How would that be calculated?

Ca(OH)2 contains one Ca2+ ion and 2 OH- ions

so 1 mole of Ca(OH)2 contains 3 moles of ions
Reply 4
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OmarEdExcel
OP
9
Original post by BobbJo
Ca(OH)2 contains one Ca2+ ion and 2 OH- ions

so 1 mole of Ca(OH)2 contains 3 moles of ions


Would that be calculated as follows :
Ca = 6.0*10^23 * 1

+

OH- = 6.0 *10^23 *2

= 1.8*10^24?
Original post by OmarEdExcel
Would that be calculated as follows :
Ca = 6.0*10^23 * 1

+

OH- = 6.0 *10^23 *2

= 1.8*10^24?

Yes
Reply 6
Avatar for OmarEdExcel
OmarEdExcel
OP
9
Original post by BobbJo
Yes


Enthalpy once again. They're tough but I am confident I'll break the chain.

http://prntscr.com/lzu1ie ( Answer is -308KJmol-1 )
Original post by OmarEdExcel
Enthalpy once again. They're tough but I am confident I'll break the chain.

http://prntscr.com/lzu1ie ( Answer is -308KJmol-1 )

Apply Hess's law
enthalpy of atomisation of silver + first ionisation energy of silver + ΔHx \Delta H_x + lattice energy = enthalpy of formation of silver nitrate
285+731+ΔHx−832=−124285 + 731 +\Delta H_x- 832 = -124
ΔHx=−308kJmol−1 \Delta H_x = -308 \text{kJmol}^{-1}

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