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2 years ago

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You are given an unknown gaseous binary compound (that is, a compound consisting of two different elements). when 10.0 g of the

compound is burned in excess oxygen, 16.3 g of water is produced. the compound has a density 1.38 times that of oxygen gas at the same conditions of temperature and pressure. give a possible identity for the unknown compound.weibo.com/u/1157067277/home?topnav=1&wvr=6

1 answer:

Luda [366]2 years ago

8 0

Answer is: a possible identity for the unknown compound is C₃H₈.

m(binary compound) = 10.0 g.

m(H₂O) = 16.3 g; mass of water.

M(O₂) = 32 g/mol; molar mass of oxygen.

M(binary compound) = 1.38 · 32 g/mol.

M(binary compound) = 44.16 g/mol.

n(binary compound) = 10 g ÷ 44.16 g/mol.

n(binary compound) = 0.225 mol; amount of substance.

n(H₂O) = 16.3 g ÷ 18 g/mol.

n(H₂O) = 0.9 mol; amount of water.

m(H₂O) : n(binary compound) = 0.9 mol ÷ 0.225 mol.

m(H₂O) : n(binary compound) = 4 : 1.

Unknown cpmpound has 4 times more hydrogen than water, it has 8 hydrogen atoms.

Second element in compound is carbon:

M(X) = 44.16 g/mol - 8 · 1.01 g/mol.

M(X) = 36.08 g/mol ÷ 3.

M(C) = 12.01 g/mol.

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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

nikitadnepr [17]

Answer:

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

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

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

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

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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