a). How the electron configurations of fluorine (F) and chlorine (Cl) are simila

Question

a). How the electron configurations of fluorine (F) and chlorine (Cl) are similar? b). How are they different? c). How do you expect the electron configurations of bromine (Br) and iodine (I) to be similar to those of fluorine and chlorine? a). Calculate the molar mass in g/mol of the following compound (ammonium dichromate): (NH_4)_2Cr_2O_7 b). Calculate the mass, in g, of the following: 0.158 mol IF_5 (iodine pentafluoride) c). Calculate the amount, in moles, of the following: 25.4 g Pb(C_2H_3O_2)_2 (lead (II) acetate) d). Calculate the percent by mass of C, H, and O in C_6 H_12 O_6 (glucose)

Explanation / Answer

F is having atomic number of 9. its electronic configuration is 1s22s22p5,

Cl is having atomic number of 17, its electronic configuration is 1s22s22p63s23p5

both suggest that there are 7 electrons in the outermost orbit of both chlorine and fluorine. since fluorine is closely bound to the nucleus ( it is in 2s orbital) and chlorine is far away from the nucleus. The electronegativity of fluorine is more than chlorine because of this.

bromine and iodine also will have same outermost electrons as compared to fluorine and chlorine.

2. formula is (NH4)2Cr2O7. Atomic weights : N= 14, H=1, Cr=52, O=16

molar mass = 2*(14+4)+2*52+6*17= 242 g/mole

3. moles of IF5= .158 moles, molar mass of IF5= 127+5*19= 222 g/mole

moles= mass/molar mass

mass= moles* molar mass= 0.158*222=35.1 gm

4. Pb(C2H3O2)2 lead acetate , atomic weights : Pb= 207 , C=12, H= 1, O=16

molar mass of lead acetate, 207+2*(2*12+3+2*16)=325 g/mole

moles of lead acetate in 25.4 gm= mass/molar mass = 25.4/325=0.078 moles

5. basis : 1 mole of glucose,

moles= mass/molar mass, 1 =mass/molar mass

mass= molar mass =180 gm

180 gm are made of 6*12 gm =72 gm of C, 12 gm of H, 6*16= 96 gm of O

mass % : C= 100*(72/180)=40, H= 100*12/180 =6.7, O= 100*(96/180)= 53.3

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