Câu 21:
Ta có: m1 + m2 = 60 (1)
\(m_{NaOH\left(10\%\right)}=10\%.m_1\left(g\right)\)
\(m_{NaOH\left(40\%\right)}=40\%.m_2\left(g\right)\)
\(\Rightarrow10\%m_1+40\%m_2=60.20\%\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_1=40\left(g\right)\\m_2=20\left(g\right)\end{matrix}\right.\)
→ Đáp án: C
Câu 22:
Gọi: \(\left\{{}\begin{matrix}m_{CuSO_4.5H_2O}=a\left(g\right)\\m_{ddCuSO_4\left(8\%\right)}=b\left(g\right)\end{matrix}\right.\) ⇒ a + b = 280 (1)
Ta có: \(n_{CuSO_4}=n_{CuSO_4.5H_2O}=\dfrac{a}{250}\left(mol\Rightarrow\right)m_{CuSO_4}=\dfrac{a}{250}.160=\dfrac{16a}{25}\left(g\right)\)
\(m_{CuSO_4\left(8\%\right)}=8\%b\left(g\right)\)
\(\Rightarrow\dfrac{16a}{25}+8\%b=280.16\%\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=40\left(g\right)\\b=240\left(g\right)\end{matrix}\right.\)
→ Đáp án: D
Câu 23:
Ta có: \(n_{HCl\left(4M\right)}=\dfrac{V}{1000}.4\left(mol\right)\)
\(n_{HCl\left(0,5M\right)}=0,4.0,5=0,2\left(mol\right)\)
\(\Rightarrow\dfrac{V}{1000}.4+0,2=\dfrac{V+400}{1000}.2\)
\(\Rightarrow V=300\left(ml\right)\)
→ Đáp án: C
Lần sau bạn đăng tách từng câu ra nhé.
Câu 16:
Ta có: \(d_{hh/H_2}=14,5\Rightarrow\overline{M}_{hh}=4,5.2=29\left(g/mol\right)\)
\(\Rightarrow\dfrac{28n_{CO}+30n_{NO}}{n_{CO}+n_{NO}}=29\)
\(\Rightarrow n_{CO}=n_{NO}\)
\(\Rightarrow\%V_{CO}=\%V_{NO}=50\%\)
→ Đáp án: A
Câu 17:
Có: \(d_{hh/H_2}=8,375\Rightarrow\overline{M}_{hh}=8,375.4=33,5\left(g/mol\right)\)
\(\Rightarrow\dfrac{44n_{N_2O}+30n_{NO}}{n_{N_2O}+n_{NO}}=33,5\)
\(\Rightarrow n_{N_2O}=\dfrac{1}{3}n_{NO}\)
\(\Rightarrow\%V_{NO}=\dfrac{n_{NO}}{n_{N_2O}+n_{NO}}.100\%=\dfrac{n_{NO}}{\dfrac{1}{3}n_{NO}+n_{NO}}.100\%=75\%\)
→ Đáp án: D
Câu 18:
Ta có: \(d_{hh/H_2}=18,5\Rightarrow\overline{M}_{hh}=18,5.2=37\left(g/mol\right)\)
\(\Rightarrow\dfrac{30n_{C_2H_6}+44n_{C_3H_8}}{n_{C_2H_6}+n_{C_3H_8}}=37\)
\(\Rightarrow n_{C_2H_6}=n_{C_3H_8}\)
\(\Rightarrow\%V_{C_2H_6}=\%V_{C_3H_8}=50\%\)
→ Đáp án: A
Câu 19:
Ta có: \(d_{hh/H_2}=25,5\Rightarrow M_{hh}=25,5.2=51\left(g/mol\right)\)
\(\Rightarrow\dfrac{44n_{C_3H_8}+58n_{C_4H_{10}}}{n_{C_3H_8}+n_{C_4H_{10}}}=51\)
\(\Rightarrow n_{C_3H_8}=n_{C_4H_{10}}\)
\(\Rightarrow\%V_{C_3H_8}=\%V_{C_4H_{10}}=50\%\)
→ Đáp án: A
Câu 20:
\(n_{HCl}=0,4.2=0,8\left(mol\right)\Rightarrow V_{HCl\left(10M\right)}=\dfrac{0,8}{10}=0,08\left(l\right)=80\left(ml\right)\)
⇒ VH2O = 400 - 80 = 320 (ml)
→ Đáp án: C
