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sửa đổi
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ancol 11 khó
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$C_{n}H_{2n+2-m}(OH)_{m}$ +mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$\Rightarrow 12n+2n+2-m+17m=38m$$\Leftrightarrow 7n+1=11m$Câu C
$C_nH_{2n+2-m}(OH)_{m} +mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_2$Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$\Rightarrow 12n+2n+2-m+17m=38m$$\Leftrightarrow 7n+1=11m$Câu C
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sửa đổi
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ancol 11 khó
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$C_{n}H_{2n+2-m}(OH)_{m}+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$\Rightarrow 12n+2n+2-m+17m=38m$$\Leftrightarrow 7n+1=11m$Câu C
$C_{n}H_{2n+2-m}(OH)_{m}$ +mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$\Rightarrow 12n+2n+2-m+17m=38m$$\Leftrightarrow 7n+1=11m$Câu C
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sửa đổi
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ancol 11 khó
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$C_{n}$H_{2n+2-m}$$(OH)_{m}$+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}Mà n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$\Rightarrow 12n+2n+2-m+17m=38m$$\Leftrightarrow 7n+1=11m$Câu C
$C_{n}H_{2n+2-m}(OH)_{m}+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$\Rightarrow 12n+2n+2-m+17m=38m$$\Leftrightarrow 7n+1=11m$Câu C
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sửa đổi
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ancol 11 khó
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$C_{n}H_{2n+2-m}(OH)_{m}+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}$Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$$\Rightarrow 12n+2n+2-m+17m=38m$$\Leftrightarrow 7n+1=11m$Câu C
$C_{n}$H_{2n+2-m}$$(OH)_{m}$+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}Mà n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$\Rightarrow 12n+2n+2-m+17m=38m$$\Leftrightarrow 7n+1=11m$Câu C
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sửa đổi
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ancol 11 khó
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$C_{n}H_{2n+2-m}(OH)_{m}$+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}Mà n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}\Rightarrow 12n+2n+2-m+17m=38m\Leftrightarrow 7n+1=11mCâu C
$C_{n}H_{2n+2-m}(OH)_{m}+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}$Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$$\Rightarrow 12n+2n+2-m+17m=38m$$\Leftrightarrow 7n+1=11m$Câu C
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sửa đổi
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ancol 11 khó
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$C_{n}H_{2n+2-m}(OH)_{m}+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}$Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$\Rightarrow 12n+2n+2-m+17m=38m\Leftrightarrow 7n+1=11mCâu C
$C_{n}H_{2n+2-m}(OH)_{m}$+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}Mà n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}\Rightarrow 12n+2n+2-m+17m=38m\Leftrightarrow 7n+1=11mCâu C
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sửa đổi
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ancol 11 khó
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C_{n}H_{2n+2-m}(OH)_{m}+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}Mà n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{x}\Rightarrow 12n+2n+2-m+17m=38m\Leftrightarrow 7n+1=11mCâu C
$C_{n}H_{2n+2-m}(OH)_{m}+mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}$Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$\Rightarrow 12n+2n+2-m+17m=38m\Leftrightarrow 7n+1=11mCâu C
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giải đáp
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ancol 11 khó
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$C_{n}H_{2n+2-m}(OH)_{m} + mNa \rightarrow C_{n}H_{2n+2-m}(ONa)_{m}+\frac{m}{2}H_{2}$Mà $n_{H_{2}}=0.1 \Rightarrow n_{C_{n}H_{2n+2-m}(OH)_{m}}=\frac{0.2}{m}$ $\Rightarrow 12n+2n+2-m+17m=38m$
$\Leftrightarrow 7n+1=11m$
Câu C
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