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Which type of hearing problem can be reduced which ordinary hearing aids

Paul [167]

Conduction deafness.

8 0

3 years ago

An ordinary flashlight battery has a potential difference of 1.2 V between its positive and negative terminals. How much work mu

Maru [420]

The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.

Given:

Potential difference, V = 1.2 V

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Calculation:

We know that the work done to transport an electron from the positive to the negative terminal is given as:

W.D = (Charge on electron)×(Potential difference)

= e × V

= (1.6 × 10⁻¹⁹ C)×(1.2 V)

= 1.92 × 10⁻¹⁹ J

Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.

Learn more about work done on a charge here:

<u>brainly.com/question/13946889</u>

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4 0

2 years ago

Sketch the resultant field pattern around the following current carrying conductors and

Novay_Z [31]

Cccccccvvvvvcjjjjjjjjjjkllk ki e

8 0

3 years ago

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

4 years ago

Calculate the speed of a train that traveled 810 kilometers in 9 hours.

Naya [18.7K]

Answer:

90 kilometers an hour

let me know if I'm wrong

8 0

3 years ago