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sertanlavr [38]
3 years ago
15

Why this chemical equation 3P4 + 10O2 = 4P2O5 + P4 is wrong?

Chemistry
1 answer:
ad-work [718]3 years ago
6 0

Because it can be balanced in an infinite number of ways so it would need to be rewritten in a simpler form.

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Biuret reagent will indicate the presence of protein in a given sample. It is also known as the Piotrowski's test. This reagent consists of copper (II) sulfate and sodium hydroxide. It detects peptide bonds by the reaction of the copper ions in an alkaline solution. The copper ions would form violet colored complexes when peptide is present in the solution. From this test, concentration can be calculated since the intensity of the color depends on the amount of peptide bonds and according to the Beer-Lambert law concentration and the absorption of light is proportional. The concentration is calculated by a spectrophotometric technique at a wavelength of 540 nm. 
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Which of the following is not a clue that a chemical reaction has taken place? *
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No colour change recorded, only change of state, hence this is a physical change - physical changes I.e. change of state and temperature are not chemical reactions.

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What is the boiling point of water at sea level ??​
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2 years ago
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 $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}

3 0
3 years ago
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