Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) mCuSO4 = 0,3.160 = 48(g)
b) \(n_{CaCO_3}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
=> mCaCO3 = 1,5.100 = 150(g)
c) \(n_{MgCl2}=\dfrac{1,5.10^{22}}{6.10^{23}}=0,025\left(mol\right)\)
=> mMgCl2 = 0,025.95 = 2,375(g)
a) \(m_{CuSO_4}=0,3.160=48\left(g\right)\)
b) \(n_{CaCO_3}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)=>m_{CaCO_3}=1,5.100=150\left(g\right)\)
c) \(n_{MgCl_2}=\dfrac{1,5.10^{22}}{6.10^{23}}=0,025\left(mol\right)=>m_{MgCl_2}=0,025.95=2,375\left(g\right)\)
e) \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=>m_{CO_2}=0,1.44=4,4\left(g\right)\)
f) \(n_{NaOH}=\dfrac{0,25.10^{24}}{6.10^{23}}=\dfrac{5}{12}\left(mol\right)=>m_{NaOH}=\dfrac{5}{12}.40=16,667\left(g\right)\)
Câu 1
\(m_{HNO_3}=0,3.63=18,9\left(g\right)\)
\(m_{CuSO_4}=1,5.160=240\left(g\right)\)
\(m_{AlCl_3}=2.133,5=267\left(g\right)\)
Câu 2
a) \(V_{N_2}=3.22,4=67,2\left(l\right)\)
\(V_{H_2}=0,45.22,4=10,08\left(l\right)\)
\(V_{O_2}=0,55.22,4=12,32\left(l\right)\)
b) \(V_{hh}=\left(0,25+0,75\right).22,4=22,4\left(l\right)\)
Làm một cái rồi tương tự nhé
\(a,\%C=\dfrac{12}{44}=27,27\%\\ \%O=100\%-27,27\%=72,73\%\)
\(a,CO\\ \%m_C=\dfrac{M_C}{M_C+M_O}.100\%=\dfrac{12}{12+16}.100\approx42,857\%\\ \Rightarrow\%m_O\approx100\%-42,857\%\approx57,143\%\\ MgCl_2\\ \%m_{Mg}=\dfrac{M_{Mg}}{M_{Mg}+2.M_{Cl}}.100\%=\dfrac{24}{24+2.35,5}.100\approx25,263\%\\ \Rightarrow\%m_{Cl}\approx100\%-25,263\%\approx74,737\%\\ C_6H_6\\ \%m_C=\dfrac{6.M_C}{6.M_C+6.M_H}.100\%=\dfrac{6.12}{6.12+6.1}.100\approx92,308\%\\ \Rightarrow\%m_H\approx100\%-92,308\%\approx7,692\%\)
a) nCO2=[(9.1023)/(6.1023)]=1,5(mol)
=> mCO2=1,5.44=66(g)
V(CO2,đktc)=1,5.22,4=33,6(l)
b) nH2=4/2=2(mol)
N(H2)=2.6.1023=12.1023(phân tử)
V(H2,đktc)=2.22,4=44,8(l)
c) N(CO2)=0,5.6.1023=3.1023(phân tử)
V(CO2,đktc)=0,5.22,4=11,2(l)
mCO2=0,5.44=22(g)
d) nN2=2,24/22,4=0,1(mol)
mN2=0,1.28=2,8(g)
N(N2)=0,1.1023.6=6.1022 (phân tử)
e) nCu=[(3,01.1023)/(6,02.1023)]=0,5(mol)
mCu=0,5.64=32(g)
Mà sao tính thể tích ta :3
\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)
a) \(m_{Na}=n.M=0,3.23=6,9\left(g\right)\)
\(m_{O_2}=n_{O_2}.M_{O_2}=0,3.32=9,6\left(g\right)\)
b) \(m_{HNO_3}=n_{HNO_3}.M_{HNO_3}=1,2.63=75,6\left(g\right)\)
\(m_{Cu}=n.M=0,5.64=32\left(g\right)\)
c) \(m_{KNO_3}=n.M=0,125=0,125.101=12,625\left(g\right)\)
\(m_{KMnO_4}=n.M=0,125.158=19,75\left(g\right)\)
\(m_{KClO_3}=n.M=0,125.122,5=15,3125\left(g\right)\)
`#3107.101107`
`3,`
a.
Khối lượng phân tử của \(\text{CaCO}_3\) là:
\(\text{M}_{\text{Ca}}+\text{M}_{\text{C}}+\text{M}_{\text{O}_3}=40+12+16\cdot3=100\left(\text{g/mol}\right)\)
b.
Khối lượng của `0,3` mol \(\text{CaCO}_3\) là:
\(\text{m}_{\text{CaCO}_3}=\text{n}_{\text{CaCO}_3}\cdot\text{M}_{\text{CaCO}_3}=0,3\cdot100=30\left(g\right)\)
Vậy:
a) `100` g/mol
b) `30` g.
$\rm a) M_{CaCO_3}=40+12+16.3=100(amu)$
$\rm b) m_{CaCO_3}=0,3.100=30(g)$
\(a.m_{CuSO_4}=n.M=0,3.160=48\left(g\right)\)
\(b.n_{CaCO_3}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\\ m_{CaCO_3}=n.M=1,5.100=150\left(g\right)\)
\(c.n_{MgCl_2}=\dfrac{1,5.10^{22}}{6.10^{23}}=0,025\left(mol\right)\\ \Rightarrow m_{MgCl_2}=n.M=0,025.95=2,375\left(g\right)\)