Question

 

Câu 16  Tính phân tử khối của phân tử các chất sau: CuO,  Ba(NO₃)₂

                                                 (Cho Cu=64, O=16,  Ba=137, N=14,)

Answer 1

\(M_{CuO}=64+16=80\left(đvc\right)\)

\(M_{Ba\left(NO_3\right)_2}=137+\left(14+16\cdot3\right)\cdot2=261\left(đvc\right)\)

Answer 2

- \(PTK_{CuO}=64+16=80\left(đvC\right)\)

- \(PTK_{Ba\left(NO_3\right)_2}=137+\left(14+16.3\right).2=261\left(đvC\right)\)

Answer 3

\(M_{CuO}=64+16=80đvc\)

\(M_{Ba\left(NO_3\right)2}=137+\left(14+16.3\right).2=261đvc\)

Answer 4

\(PTK_{CuO}=64+16=80đvC\)

\(PTK_{Ba\left(NO_3\right)_2}=137+\left(14+16.3\right).2=261đvC\)

Answer 5

CuO=64.1+16.1=80 dVc

Ba(NO₃)₂= 137.1+ (14.1+16.3).2= 261 dVc