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3 years ago

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Chemistry

1 answer:

Dovator [93]3 years ago

3 0

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

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For the reactants,

The oxidation number of hydrogen = +1
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For the products,

The oxidation number of hydrogen = +1
The oxidation number of oxygen = -2
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Here, arsenic (+5 to +3) and carbon (+3 to +4) are the only oxidation numbers changing.

Note that an increase in oxidation number means electrons are lost. Thus oxidation is occurring, and a decrease in oxidation number means electrons are being gained, and thus reduction is occurring.

Also, the compound that contains the element being oxidized is the reducing agent, and the compound that contains the element being reduced is the oxidizing agent.

So, the answers are:

name of the element oxidized: Carbon

name of the element reduced: Arsenic

formula of the oxidizing agent: $\text{H}_{3}\text{AsO}_{4}$

formula of the reducing agent: $\text{H}_{2}\text{C}_{2}\text{O}_{4}$

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2 years ago

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This is therefore the reason why it was chosen as the most appropriate choice.

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1 year ago

When hydrogen gas reacts with oxygen gas under certain conditions to produce liquid water, 572.0 kJ of heat energy is released.

frozen [14]

Answer : This reaction is an exothermic reaction.

Explanation :

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In the endothermic reaction, the energy of reactant are less than the energy of product.

Exothermic reaction : It is defined as the chemical reaction in which the energy is released into the surrounding.

In the exothermic reaction, the energy of reactant are more than the energy of product.

Enthalpy of reaction : It is the difference between the energy of product and the reactant. It is represented as $\Delta H$ .

The balanced chemical reaction will be:

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3 years ago

The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding

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Answer:

$Binding\ energy=43.43\times 10^{-20}\ J$

Explanation:

Using the expression for the photoelectric effect as:

$E=h\nu_0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

$\nu_0=\frac {c}{\lambda_0}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light being bombarded

Given, $\lambda=4.00\times 10^{-7}\ m$

$\frac {hc}{\lambda_0}$ is the binding energy or threshold energy

$\frac {1}{2}\times m\times v^2$ is the kinetic energy of the electron emitted. = $6.26\times 10^{-20}\ J$

Thus, applying values as:

$\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy$

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

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3 years ago