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2 years ago

Study the reactions.

Chemistry

1 answer:

yulyashka [42]2 years ago

3 0

Answer:

$\displaystyle \Delta H = -2426\text{ kJ}$

Explanation:

To find the enthalpy change of the target reaction, we can use Hess's Law.

Reversing the third reaction yields:

$\displaystyle \text{C$_3$H$_8$(g)} \longrightarrow 3\text{C(s)} + 4\text{H$_2$(g)}\;\;\;\;\;\Delta H = -106\text{ kJ}$

Multiplying the first reaction by three yields:

$\displaystyle 3\text{C(s)} + 3\text{O$_2$(g)} \longrightarrow 3\text{CO$_2$}(g)}\;\;\;\;\; \Delta H = -1.18\times 10^3\text{ kJ}$

Multiplying the second reaction by four yields:

$\displaystyle 4\text{H$_2$(g)} + 2\text{O$_2$(g)} \longrightarrow 4\text{H$_2$O($\ell$)} \;\;\;\;\; \Delta H = -1.14\times 10^3 \text{ kJ}$

Adding all equations yield:

$\displaystyle \text{C$_3$H$_8$(g)} + 5\text{O$_2$(g)} \longrightarrow 3\text{CO$_2$(g)} + 4\text{H$_2$O($\ell$)} \;\;\;\;\; \Delta H = -2426\text{ kJ}$

Hence, the enthalpy change of the target reaction is -2426 kJ.

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Answer:

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Explanation:

Given that:

The work function of the magnesium = 2.3 eV

Energy in eV can be converted to energy in J as:

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Using the equation for photoelectric effect as:

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Applying the equation as:

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Where,

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

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

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

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=260\ nm=260\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}$

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$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}$

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