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\(1dvC=\dfrac{1}{12}.m_C=\dfrac{1,9926.10^{-23}}{12}=0,16605.10^{-23}\left(g\right)\)
\(NTK_{Cu}=64dvC=64.0,16605.10^{-23}=10,6272.10^{-23}\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(m_K=39\cdot1.66\cdot10^{-24}=6.474\cdot10^{-23}\left(g\right)\)
\(m_{Zn}=65\cdot1.66\cdot10^{-24}=1.079\cdot10^{-22}\left(g\right)\)
\(m_{Cu}=64\cdot1.66\cdot10^{-24}=1.0624\cdot10^{-22}\left(g\right)\)
\(m_{Mg}=24\cdot1.66\cdot10^{-24}=3.984\cdot10^{-23}\left(g\right)\)
b.
\(m_{Na_2O}=62\cdot1.66\cdot10^{-24}=1.0292\cdot10^{-22}\left(g\right)\)
\(m_{CaO}=56\cdot1.66\cdot10^{-24}=9.296\cdot10^{-23}\left(g\right)\)
\(m_{FeCl_2}=127\cdot1.66\cdot10^{-24}=2.1082\cdot10^{-22}\left(g\right)\)
\(m_{Al_2O_3}=102\cdot1.66\cdot10^{-24}=1.6932\cdot10^{-22}\left(g\right)\)
a) Khối lượng tính bằng gam của:
\(m_K=0,16605.10^{-23}.39=6,47595.10^{-23}\left(g\right)\)
\(m_{Zn}=0,16605.10^{-23}.65=10,79325.10^{-23}\left(g\right)\\ m_{Cu}=0,16605.10^{-23}.64=10,6272.10^{-23}\left(g\right)\\ m_{Mg}=0,16605.10^{-23}.24=3,9852.10^{-23}\left(g\right)\)
b) Khối lượng tính bằng gam của các phân tử:
\(m_{Na_2O}=62.0,16605.10^{-23}=10,2951.10^{-23}\left(g\right)\\ m_{CaO}=56.0,16605.10^{-23}=9,2988.10^{-23}\left(g\right)\\ m_{FeCl_2}=127.0,16605.10^{-23}=21,08835.10^{-23}\left(g\right)\\ m_{Al_2O_3}=102.0,16605.10^{-23}=16,9371.10^{-23}\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{Ca}=m_{1\text{đ}vC}\times NTK_{Ca}=0,166\times10^{-23}g\times40=6,64\times10^{-23}g\)
\(m_{Cu}=m_{1\text{đ}vC}\times NTK_{Cu}=0,166\times10^{-23}g\times64=10,624\times10^{-23}g\)
\(m_S=m_{1\text{đ}vC}\times NTK_S=0,166\times10^{-23}g\times32=5,312\times10^{-23}g\)
\(m_{Ba}=m_{1\text{đ}vC}\times NTK_{Ba}=0,166\times10^{-23}g\times137=22,742\times10^{-23}g\)
\(m_{Ag}=m_{1\text{đ}vC}\times NTK_{Ag}=0,166\times10^{-23}g\times27=4,482\times10^{-23}g\)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
Khối lượng nguyên tử Cu = \(\dfrac{64}{12}.1,9926.10^{-23}=10,63.10^{-23}\left(g\right)\)
=> A
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
$M_X = 2X = 5.16 = 80 \Rightarrow X = 40$
Vậy X là canxi, KHHH : Ca
b)
$m_{Ca} = 40.1,66.10^{-24} = 66,4.10^{-24}(gam)$
c)
$m_{5Ca} = 5.66,4.10^{-24} = 332.10^{-24}(gam)$
![](https://rs.olm.vn/images/avt/0.png?1311)
Khối lượng nguyên tử Đồng (Cu) là
\(\dfrac{\text{64.1 , 9926.10}^{-23}}{12}\) = \(1.0627.10\)-22 \((gam)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{1\text{đ}vC}=\frac{1}{12}\times m_C=\frac{1}{12}\times1,9926\times10^{-23}g=0,166\times10^{-23}g\)
\(m_{gam}\left(P\right)=m_{1\text{đ}vC}\times NTK\left(P\right)=0,166\times10^{-23}g\times31\approx5,15\times10^{-23}g\)
nguyên tử khối của photpho là 31 (1)
khối lượng tính bằng gam của 1 đơn vị cacbon là :
(1,9926 * 10-23) : 12 = 1,66 * 10-24(g)
=> khối lượng tính bằng gam của nguyên tử phốt pho
là : 1,66 * 10-24 * 31 = 5,1*10-23 (g)
5Cu= 5.64=320
m Cu=320.1,66.10-24=5,312.10-22g