Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) mS=32.1=32(g)
mC=0,25.12=3(g)
mCl2=0,2.71=14,2(g)
mMg=0,6.24=14,4(g)
b) mN=0,1.28=2,8(g)
mCu=0,25.64=16(g)
c) mNaCl=58,5.1,75=102,375(g)
mC6H12O6=0,25.180=45(g)
a) \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)
\(n_{CaCO_3}=\dfrac{25}{100}=0,25\left(mol\right)\)
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(n_{...}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)
b)
\(m_{ZnSO_4}=0,25.161=40,25\left(g\right)\)
\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
\(m_{Cu}=0,3.64=19,2\left(g\right)\)
\(m_{Fe_2\left(SO_4\right)_3}=0,35.400=140\left(g\right)\)
d) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
\(V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)
\(V_{SO_2}=0,3.22,4=6,72\left(l\right)\)
Câu 1:
a) \(m_{Fe_2\left(SO_4\right)_3}=0,15.400=60\left(g\right)\)
b) \(m_{MgCl_2}=0,05.95=4,75\left(g\right)\)
c) \(m_{H_2}=0,2.2=0,4\left(g\right)\)
d) \(n_{N_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow m_{N_2}=0,2.28=5,6\left(g\right)\)
e) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\Rightarrow m_{O_2}=0,3.32=9,6\left(g\right)\)
Câu 2:
a) \(V_{NO_2}=0,25.22,4=5,6\left(mol\right)\)
b) \(V_{CO_2}=0,3.22,4=6,72\left(mol\right)\)
c) \(n_{Cl_2}=\dfrac{3,55}{35,5}=0,1\left(mol\right)\Rightarrow V_{Cl_2}=0,1.22,4=2,24\left(l\right)\)
d) \(n_{N_2O}=\dfrac{1,32}{44}=0,03\left(mol\right)\Rightarrow V_{N_2O}=0,03.22,4=0,672\left(l\right)\)
N phân tử = 1 mol phân tử
\(\Rightarrow n_{O2}=1mol;n_{N_2}=2mol;n_{CO_2}=1,5mol\)
\(\Rightarrow m_{hh}=1.32+2.28+1,5.44=154g\)
b. \(m_{hh}=0,1.56+0,2.64+0,3.65+0,25.27=44,65g\)
c. \(n_{O_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)
\(n_{HCl}=\dfrac{6,72}{22,4}=0,3mol\)
\(n_{CO_2}=\dfrac{0,56}{22,4}=0,025mol\)
\(\Rightarrow m_{hh}=0,1.32+0,05.2+0,3.36,5+0,025.44=15,35g\)
a, \(\overline{M}=\dfrac{0,1.44+0,2.28}{0,1+0,2}\approx33,33\left(g/mol\right)\)
b, \(\overline{M}=\dfrac{0,2.28+0,3.2}{0,2+0,3}=12,4\left(g/mol\right)\)
c, \(\overline{M}=\dfrac{0,1.28+0,2.30+0,2.44}{0,1+0,2+0,2}=35,2\left(g/mol\right)\)
d, \(\overline{M}=\dfrac{0,2.56+0,1.24+0,1.27}{0,2+0,1+0,1}=40,75\left(g/mol\right)\)
Bài làm
* \(m_{ZnSO4}=n.M=0,25.\left(65+32+16.4\right)=0,25.161=40,25\left(g\right)\)
* \(m_{AlCl3}=n.M=0,2.\left(27+35,5.3\right)=0,2.133,5=26,7\left(g\right)\)
* \(m_{Cu}=n.M=0,3.64=19,2\left(g\right)\)
* \(m_{Ca\left(OH\right)2}=n.M=0,15.\left[40+\left(16+1\right).2\right]=0,15.74=11,1\left(g\right)\)
* \(m_{Fe2\left(SO4\right)3}=n.M=0,35.\left[56+\left(32+16.4\right).3\right]=0,35.344=120,4\left(g\right)\)
# Học tốt #
\(m_{H_2S}=2.34=68\left(g\right)\)
\(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)
\(m_{CuO}=0,5.80=40\left(g\right)\)
áp dụng công thức : m = n.M
Ta có : \(m_{H2S}\) = 2.34 = 68 (g)
\(m_{H2SO4}\) = 0,25 .98= 24,5 (g)
\(m_{CuO}\) = 0,5 . 80 = 40 (g)
a) mFe = 0,3.56 = 16,8 (g)
b) mMg = 0,4.24 = 9,6 (g)
c) mH2O = 0,2.18 = 3,6 (g)
d) mH2SO4 = 0,25.98 = 24,5(g)
a: \(m=M\cdot n=56\cdot0.3=16.8\left(g\right)\)
b: \(m=M\cdot n=24\cdot0.4=9.6\left(g\right)\)