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3 years ago

I don’t really understand, if anybody can help I’ll really appreciate it ! Thank you.

Chemistry

1 answer:

Alex_Xolod [135]3 years ago

3 0

Answer:

175

Explanation:

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zhuklara [117]

Answer:

Explanation:

I think it is C

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3 years ago

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici

Brut [27]

Question:

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

Answer:

The mean activity coefficient for HCl solution is 0.78.

Explanation:

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

$E=E_{0}-0.0514 \mathrm{V} \ln \gamma$

As we know that $E_{0}$ = 0.22 V and E = 0.342 V, the equation will become

$0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma$

$0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)$

$0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)$

$\frac{0.12}{0.0514}=-\ln (0.124 \gamma)$

$-2.3346=\ln (0.124 \gamma)$

$e^{-2.3346}=0.124 \gamma$

$0.0968=0.124 \gamma$

$\gamma=\frac{0.0968}{0.124}=0.78$

So, the mean activity coefficient is 0.78.

6 0

3 years ago

A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('

polet [3.4K]

Answer:

$a=5.65cm$

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

$Q_{Pt}=-Q_{Deu}$

We can represent the heats in terms of mass, heat capacities and temperatures:

$m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})$

Thus, we solve for the mass of platinum:

$m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g$