Question

Câu 12. Tính khối lượng phân tử theo đvC của các phân tử sau. Cho biết chất nào là đơn chất, chất nào là hợp chất. a) C, Cl₂, KOH, H₂SO4, Fe₂(CO3)3. b) BaSO4, O2, Ca(OH)2, Fe. a c) HCI, NO, Br2, K, NH3. d) CH;OH, CH4, O3, BaO.

Answer 1

Dạng này em tính phân tử khối, nguyên tử khối rồi nhân với 0,16605.10^-23 (g)

Answer 2

Trả lời:

\(a)\)

\(m_C=1,6605.10^{-24}.12=1,9926.10^{-23}\left(g\right)\)

\(m_{Cl}=1,6605.10^{-24}.35,5=5,894775.10^{-23}\left(g\right)\)

\(m_{KOH}=1,6605.10^{-24}.\left(39+16+1\right)=9,2988.10^{-23}\left(g\right)\)

\(m_{H2SO4}=1,6605.10^{-24}.\left(2+32+4.16\right)=1,62729.10^{-22}\left(g\right)\)

\(m_{Fe2\left(CO3\right)3}=1,6605.10^{-24}.\left(2.56+\left(12+3.16\right).3\right)=4,84866.10^{-22}\left(g\right)\)

+) Đơn chất: \(C,Cl.\)

+) Hợp chất: \(KOH,H_2SO_4,Fe_2\left(CO_3\right)_3.\)

\(b)\)

\(m_{BaSO4}=1,6605.10^{-24}.\left(137+32+4.16\right)=3,868965.10^{-22}\left(g\right)\)

\(m_{O2}=1,6605.10^{-24}.\left(2.16\right)=5,3136.10^{-23}\left(g\right)\)

\(m_{Ca\left(OH\right)2}=1,6605.10^{-24}.\left(40+\left(16+1\right).2\right)=1,22877.10^{-22}\left(g\right)\)

\(m_{Fe}=1,6605.10^{-24}.56=9,2988.10^{-23}\left(g\right)\)

+) Đơn chất: \(O_2,Fe.\)

+) Hợp chất: \(BaSO_4,Ca\left(OH\right)_2.\)

\(c)\)

\(m_{HCl}=1,6605.10^{-24}.\left(1+35,5\right)=6,060825.10^{-23}\left(g\right)\)

\(m_{NO}=1,6605.10^{-24}.\left(14+16\right)=4,9815.10^{-23}\left(g\right)\)

\(m_{Br2}=1,6605.10^{-24}.\left(2.80\right)=2,6568.10^{-22}\left(g\right)\)

\(m_K=1,6605.10^{-24}.39=6,47595.10^{-23}\left(g\right)\)

\(m_{NH3}=1,6605.10^{-24}.\left(14+3.1\right)=2,82285.10^{-23}\left(g\right)\)

+) Đơn chất: \(Br_2,K.\)

+) Hợp chất: \(HCl,NO,NH_3.\)

\(d)\)

\(m_{C6H5OH}=1,6605.10^{-24}.\left(12.6+5.1+16+1\right)=1,56087.10^{-22}\left(g\right)\)\(m_{CH4}=1,6605.10^{-24}.\left(12+4.1\right)=2,6568.10^{-23}\left(g\right)\)

\(m_{O3}=1,6605.10^{-24}.\left(3.16\right)=7,9704.10^{-23}\left(g\right)\)

\(m_{BaO}=1,6605.10^{-24}.\left(137+16\right)=2,540565.10^{-22}\left(g\right)\)

+) Đơn chất: \(O_3\)

+) Hợp chất: \(C_6H_5OH,CH_4,BaO.\)

Answer 3

Bài làm:

$a)$

$m_C=1,6605.10^{-24}.12=1,9926.10^{-23}\left(g\right)$

$m_{Cl}=1,6605.10^{-24}.35,5=5,894775.10^{-23}\left(g\right)$

$m_{KOH}=1,6605.10^{-24}.\left(39+16+1\right)=9,2988.10^{-23}\left(g\right)$

$m_{H2SO4}=1,6605.10^{-24}.\left(2+32+4.16\right)=1,62729.10^{-22}\left(g\right)$

$m_{Fe2\left(CO3\right)3}=1,6605.10^{-24}.\left(2.56+\left(12+3.16\right).3\right)=4,84866.10^{-22}\left(g\right)$

+) Đơn chất: $C,Cl.$

+) Hợp chất: $KOH,H_2SO_4,Fe_2\left(CO_3\right)_3.$

$b)$

$m_{BaSO4}=1,6605.10^{-24}.\left(137+32+4.16\right)=3,868965.10^{-22}\left(g\right)$

$m_{O2}=1,6605.10^{-24}.\left(2.16\right)=5,3136.10^{-23}\left(g\right)$

$m_{Ca\left(OH\right)2}=1,6605.10^{-24}.\left(40+\left(16+1\right).2\right)=1,22877.10^{-22}\left(g\right)$

$m_{Fe}=1,6605.10^{-24}.56=9,2988.10^{-23}\left(g\right)$

+) Đơn chất: $O_2,Fe.$

+) Hợp chất: $BaSO_4,Ca\left(OH\right)_2.$

$c)$

$m_{HCl}=1,6605.10^{-24}.\left(1+35,5\right)=6,060825.10^{-23}\left(g\right)$

$m_{NO}=1,6605.10^{-24}.\left(14+16\right)=4,9815.10^{-23}\left(g\right)$

$m_{Br2}=1,6605.10^{-24}.\left(2.80\right)=2,6568.10^{-22}\left(g\right)$

$m_K=1,6605.10^{-24}.39=6,47595.10^{-23}\left(g\right)$

$m_{NH3}=1,6605.10^{-24}.\left(14+3.1\right)=2,82285.10^{-23}\left(g\right)$

+) Đơn chất: $Br_2,K.$

+) Hợp chất: $HCl,NO,NH_3.$

$d)$

$m_{C6H5OH}=1,6605.10^{-24}.\left(12.6+5.1+16+1\right)=1,56087.10^{-22}\left(g\right)\)\(m_{CH4}=1,6605.10^{-24}.\left(12+4.1\right)=2,6568.10^{-23}\left(g\right)$

$m_{O3}=1,6605.10^{-24}.\left(3.16\right)=7,9704.10^{-23}\left(g\right)$

$m_{BaO}=1,6605.10^{-24}.\left(137+16\right)=2,540565.10^{-22}\left(g\right)$

+) Đơn chất: $O_3$

+) Hợp chất: $C_6H_5OH,CH_4,BaO.$