Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`n_{FeSO_4} = 0,25(mol)`
\(n_{FeSO_4}=\dfrac{13,2.10^{23}}{6.10^{23}}=2,2\left(mol\right)\)
\(n_{NO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_A=n_{Al}+n_{Cu}=0,22+0,25=0,47\left(mol\right)\)
\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right);n_{N_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \Rightarrow n_B=n_{O_2}+n_{N_2}=0,5+0,6=1,1\left(mol\right)\)
\(n_{Fe}=\dfrac{15.10^{23}}{6.10^{23}}=2,5\left(mol\right)\\ \Rightarrow n_C=n_{Cu}+n_{Fe}=0,25+2,5=2,75\left(mol\right)\\ n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right);n_{N_2}=\dfrac{2,7.10^{23}}{6.10^{23}}=0,45\left(mol\right)\\ \Rightarrow n_D=n_{O_2}+n_{CO_2}+n_{N_2}=0,25+0,5+0,45=1,2\left(mol\right)\)
a.
\(m_{Al}=0.5\cdot27=13.5\left(g\right)\)
\(m_{CO_2}=\dfrac{6.72}{22.4}\cdot44=13.2\left(g\right)\)
\(m_{N_2}=\dfrac{5.6}{22.4}\cdot28=7\left(g\right)\)
\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)
b.
\(m_{hh}=\dfrac{3.36}{22.4}\cdot2+\dfrac{5.6}{22.4}\cdot28+0.2\cdot44=16.1\left(g\right)\)
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. \(n_{O_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)
\(n_{HCl}=\dfrac{6,72}{22,4}=0,3mol\)
\(n_{CO_2}=\dfrac{0,56}{22,4}=0,025mol\)
\(\Rightarrow m_{hh}=0,1.32+0,05.2+0,3.36,5+0,025.44=15,35g\)
1)
Số nguyên tử Al là 1,5.6.1023 = 9.1023 nguyên tử
Số phân tử H2 là 0,5.6.1023 = 3.1023 phân tử
Số phân tử NaCl là 0,25.6.1023 = 1,5.1023 phân tử
Số phân tử H2O là 0,05.6.1023 = 0,3.1023 phân tử
2)
$V_{CO_2} = 1.22,4 = 22,4(lít)$
$V_{H_2} = 2.22,4 = 44,8(lít)$
$V_{O_2} = 1,5.22,4 = 33,6(lít)$
$V_{hh} = (0,25 + 1,25).22,4 = 33,6(lít)$
1. Tính số nguyên tử hoặc phân tử trong mỗi lượng chất sau: 1,5 mol Al; 0,5 mol H2; 0,25 mol NaCl; 0,05 mol H2O.
1,5 mol Al có : \(1,5.6.10^{23}=9.10^{23}\) nguyên tử
0,5 mol H2 có : \(0,5.6.10^{23}=3.10^{23}\)phân tử
0,25 mol NaCl: \(0,25.6.10^{23}=1,5.10^{23}\) phân tử
0,05 mol H2O có : \(0,05.6.10^{23}=0,3.10^{23}\) phân tử
mMg = 0,5.24 = 12 gam
VSO2 = n.22,4 = 0,25.22,4 = 5,6 lít
nN2 = \(\dfrac{16,8}{22,4}\)= 0,75 mol , nO2 = \(\dfrac{5,6}{22,4}\)= 0,25 mol
=> m(N2 + O2 ) = 0,75.28 + 0,25.32 = 29 gam
\(a,m_{CaSO_4}=136.0,25=34\left(g\right)\\ b,n_{Cu_2O}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ m_{Cu_2O}=0,5.144=72\left(g\right)\\ c,n_{NH_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ m_{NH_3}=17.0,3=5,1\left(g\right)\\ d,m_{C_4H_{10}}=0,17.58=9,86\left(g\right)\\ e,n_{Cu\left(OH\right)_2}=\dfrac{4,5.10^{25}}{6.10^{23}}=75\left(mol\right)\\ m_{Cu\left(OH\right)_2}=98.75=7350\left(g\right)\\ g,m_{MgO}=0,48.40=19,2\left(g\right)\\ h,n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ m_{CO_2}=44.0,15=6,6\left(g\right)\\ i,m_{Al\left(OH\right)_3}=78.0,25=19,5\left(g\right)\\\)
Các câu còn lại em làm tương tự nha!
\(a_1.m_{CaCl_2}=n.M=0,25.111=27,75\left(g\right)\\ m_{Cu\left(OH\right)_2}=n.M=0,5.98=49\left(g\right)\\ \Rightarrow m_{hh}=m_{CaCl_2}+m_{Cu\left(OH\right)_2}=27,75+49=76,75\left(g\right)\)
\(b_1.n_{N_2}=\dfrac{V_{\left(dktc\right)}}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{O_2}=\dfrac{V_{\left(dktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ m_{N_2}=n.M=0,25.28=7\left(g\right)\\ m_{O_2}=n.M=0,15.32=4,8\left(g\right)\\ \Rightarrow m_{hh}=m_{N_2}+m_{O_2}=7+4,8=11,8\left(g\right)\)
\(c_1.n_{H_2O}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ n_{CO}=\dfrac{2,4.10^{23}}{6.10^{23}}=0,4\left(mol\right)\\ m_{H_2O}=n.M=0,5.18=9\left(g\right)\\ m_{CO}=n.M=0,4.28=11,2\left(g\right)\\ m_{hh}=m_{H_2O}+m_{CO}=9+11,2=20,2\left(g\right)\)
\(a_2.n_{hh}=n_{Cl_2}+n_{N_2}=0,25+0,3=0,55\left(mol\right)\\ V_{hh\left(dktc\right)}=n.22,4=0,55.22,4=12,32\left(l\right)\)
\(n_{SO_2}=\dfrac{m}{M}=\dfrac{3,2}{64}=0,05\left(mol\right)\\ n_{N_2}=\dfrac{9,8}{28}=0,35\left(mol\right)\\ n_{hh}=n_{SO_2}+n_{N_2}=0,05+0,35=0,4\left(mol\right)\\ V_{hh}=n.22,4=0,4.22,4=8,96\left(l\right)\)
a) mFeSO4= 0,25.152=38(g)
b) mFeSO4= \(\dfrac{13,2.10^{23}}{6.10^{23}}.152=334,4\left(g\right)\)
c) mNO2= \(\dfrac{8,96}{22,4}.46=18,4\left(g\right)\)
d) mA= 27.0,22+64.0,25=21,94(g)
e) mB= \(\dfrac{11,2}{22,4}.32+\dfrac{13,44}{22,4}.28=32,8\left(g\right)\)
g) mC= \(64.0,25+\dfrac{15.10^{23}}{6.10^{23}}.56=156\left(g\right)\)
h) mD= \(0,25.32+\dfrac{11,2}{22,4}.44+\dfrac{2,7.10^{23}}{6.10^{23}}.28=42,6\left(g\right)\)
hơi muộn nha<3