Question

1- Calculate the following:

A- Number of moles in 10 gram of AlBr3.

B- Number of Atoms in 10 gram of Al(OH)3.

C- Number of molecules in 10 gram of Cr(NO3)5.

D- Number of grams in 10 10 atoms of AlBr3.

2- How many gram of Oxygen is in 10 gram of Zn(OH)2?

3- Balance the following chemical equations:

A- ​O2​​​--------------->​​O3

B-​Na+​+​Al(OH​)3​-------------->

C-​KI​+​Mg(NO3)2​--------------->​

D-​FeCl3​+​Ca3(PO4)2​--------------->​

E- C4H12​+​O2​---------------->​CO2​+​H2O

4- Find Oxidized –Reduced and number of electrons:

a) 3Ba​+​N2​---------->​Ba3N2

OX:

RE:

Electrons:

b) 3Ba​+​2AlN​---------->​Ba3N2​+​2Al

OX:

RE:

Electrons:

Answer 1

266.7 - 10 gm Al Brz. molan mass of Al B&z= 266-7 9/mol a moles of Al Br 3 = 10 = 0.0375 moles 10 gm Allohla A molar mass of AllOH)3 = 78 olmal a molles of Al (OH)3 = + = 0-138 miles 1 mol has 6.023X1023 units a holof Allotis malecules = 0.128 X 6.0 23x1023 = 7.72 X1622 molecules . I molecule of Al(OH)3 has 7 atoms (1 Al, 30, 3H) no of atoms = 7x772X1022 • a (no. of atony = 5.41 x 1023 atoms) 10 gm cr (noils molan mass of co(003) = 362 91ml • a moles of (8(903)5 = 10 y = 0.0276 moles - a (no of co(NO3)s molecules = 0.0276X6-023 61023 - = 1.66X1022 molecules - 1 1 molecule of Albr3 has 4 anons - a ido atorms will have 1000 = 2.5x100 molecules 2 x 1 mol has 600 23 1823 molecules - a moles of AlB8= 25x10J = 415x15ls males 6-023X1023 : molar mass of Alboz = 26607 9/mo fes Scanmaswalf Al Br 3 = 268 +7+ 4- 15x1815 = 111x151 gm CamScanner: