Question

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1- Calculate the following: A- Number of moles in 10 gram of Cr(NO3)s. B- Number of Atoms in 2.5 gram of Cr(NO3)s. C- Number of molecules in 100 gram of Fe(NO3)3. D- Number of grams in 10 20 atoms of MgBr2.

Answer 1

1A) 1 mole of Cr(NO3)3 is 238 g

10 g Cr(NO3)3 = 10/238

= 4.2×10^-2 moles

B) 238 g Cr(NO3)3 contain 6.022×10²³ molecules

238 g Cr(NO3)3 contain 13×6.022×10²³ atoms.

2.5 g Cr(NO3)3 contain 13×6.022×10²³/2.5 atoms

= 31.3×10²³ atoms.

C) 241.86 g Fe(NO3)3 contain 6.022×10²³ molecules

100 g Fe(NO3)3 contain 6.022×100×10²³/241.86 molecules

= 2.49×10²³ molecules

D) 184 g MgBr2 present in 3× 6.022×10²³ atoms

10²⁰ atoms are present in 184×10²⁰/3×6.022×10²³ atoms

= 0.01 atoms