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3 years ago

28 Which compound yields H + ions as the only positive ions in an aqueous solution?

Chemistry

1 answer:

MrMuchimi3 years ago

4 0

Answer: (4) $CH_3COOH$

Explanation:-

According to Arrhenius concept, a base is defined as a substance which donates hydroxide ions $(OH^-)$ when dissolved in water and an acid is defined as a substance which donates hydronium ions $(H_3O^+)$ in water.

1. $KOH$ is a strong electrolyte and thus dissociate in water. $KOH\rightarrow K^++OH^-$

2. $CH_3OH$ is a non electrolyte and thus do no dissociate in water.

$CH_3COOH\rightarrow no reaction$

3. $NaOH$ is a strong electrolyte and thus dissociate in water.

$NaOH\rightarrow Na^++OH^-$

4. $CH_3COOH$ is a weak electrolyte and thus partially dissociate in water.

$CH_3COOH\rightleftharpoons CH_3COO^-+H^+$

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A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

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Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

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Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

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For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

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