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3 years ago

Class material don't interact please

Chemistry

1 answer:

bagirrra123 [75]3 years ago

8 0

A 0.00143 M concentration of MnO4^- is not a reasonable solution .

<h3>Number of moles of carbonate</h3>

The ions left in solution are Na^+ and NO3^-

Number of moles of calcium nitrate = 100/1000 L × 1 = 0.1 moles

Since;

1 mole of sodium carbonate reacts with 1 mole of calcium nitrate then 0.1 moles of sodium carbonate were used.

<h3>Conductivity of filtrate</h3>

The claim of the student that the concentration of sodium carbonate is too low is wrong because the value was calculated from concentration and volume of calcium nitrate and not using the precipitate. If the filtrate is tested for conductivity, it will be found to conduct electricity because it contains sodium and NO3 ions.

2) In the reaction as shown, the MnO4^- ion was reduced.

The initial volume is 3.4 mL while the final volume is 29.6 mL.

Number of moles of MnO4^- ion = (29.6 mL - 3.4 mL)/1000 × 0.0235 M = 0.0006157 moles

<h3>The calculations are performed as follows</h3>

If 2 moles of MnO4^- reacted with 5 moles of acid

0.0006157 moles of MnO4^- reacted with 0.0006157 moles × 5 moles/ 2 moles

= 0.0015 moles

In this case, number of moles of acid = 0.139 g/90 g/mol = 0.0015 moles

Number of moles of MnO4^- = 0.00143 M × (29.6 mL - 3.4 mL)/1000

= 0.000037 moles

If 2 moles of MnO4^- reacts with 5 moles of acid

0.000037 moles of MnO4^- reacts with 0.000037 moles × 5 moles/ 2 moles

= 0.000093 moles

Hence, this is not a reasonable amount of solution.

Learn more about MnO4^- : brainly.com/question/10887629

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a metal spoon left in boiling water

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3 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 $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$