Question

[Trần Jungkook]

tính khối lượng bằng gam của nguyên tử Ca ; Cu ; S ; Ba ; Ag

Answer 1

[Phương An]

\(m_{Ca}=m_{1\text{đ}vC}\times NTK_{Ca}=0,166\times10^{-23}g\times40=6,64\times10^{-23}g\)

\(m_{Cu}=m_{1\text{đ}vC}\times NTK_{Cu}=0,166\times10^{-23}g\times64=10,624\times10^{-23}g\)

\(m_S=m_{1\text{đ}vC}\times NTK_S=0,166\times10^{-23}g\times32=5,312\times10^{-23}g\)

\(m_{Ba}=m_{1\text{đ}vC}\times NTK_{Ba}=0,166\times10^{-23}g\times137=22,742\times10^{-23}g\)

\(m_{Ag}=m_{1\text{đ}vC}\times NTK_{Ag}=0,166\times10^{-23}g\times27=4,482\times10^{-23}g\)

Answer 2

[Huy Giang Pham Huy]

có Ca=40đvC

nên Cu=40\(\times\)0,16605\(\times\)10^-23=6,642\(\times\)10^-23(gam)

có S=32đvC

nên S=\(32\times0,16605\times10^{-23}=5,3136\times10^{-23}\)(gam)

có Ba = 137 đvC

nên Ba=137\(\times\)0,16605\(\times\)10^-23=22,74885\(\times\)10^-23(gam)

có Ag=108đvC

nên Ag=\(108\times0,16605\times10^{-23}=17,9334\times10^{-23}\)(gam)