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3 years ago

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Green plants absorbs sunlight to power photosynthesis, the chemical synthesis of food from water and carbon dioxide. The compoun

d responsible for light absorption and the color of plants, chlorophyll, strongly absorbs light with a wavelength of 453.nm. Calculate the frequency of this light.

1 answer:

Serggg [28]3 years ago

3 0

Answer: The frequency of this light is $6.62\times 10^{14}s^{-1}$

Explanation:

To calculate the wavelength of light, we use the equation:

$\lambda=\frac{c}{\nu}$

where,

$\lambda$ = wavelength of the light = $453nm=453\times 10^{-9}m$

c = speed of light = $3.0\times 10^8m/s$

$\nu$ = frequency of light = ?

$\nu=\frac{3.0\times 10^8m/s}{453\times 10^{-9}m}=6.62\times 10^{14}s^{-1}$

The frequency of this light is $6.62\times 10^{14}s^{-1}$

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The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching it

ollegr [7]

Explanation:

It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.

This means that volume occupied by 2 atoms is equal to volume of the unit cell.

So, according to the volume density

$5 \times 10^{26} atoms = 1 [tex]m^{3}$

2 atoms = $\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms$

= $4 \times 10^{-27} m^{3}$

Formula for volume of a cube is $a^{3}$ . Therefore,

Volume of the cube = $4 \times 10^{-27} m^{3}$

As lattice constant (a) = $(4 \times 10^{-27} m^{3})^{\frac{1}{3}}$

= $1.59 \times 10^{-9} m$

Therefore, the value of lattice constant is $1.59 \times 10^{-9} m$ .

And, for bcc unit cell the value of radius is as follows.

r = $\frac{\sqrt{3}}{4}a$

Hence, effective radius of the atom is calculated as follows.

r = $\frac{\sqrt{3}}{4}a$

= $\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m$

= $6.9 \times 10^{-10} m$

Hence, the value of effective radius of the atom is $6.9 \times 10^{-10} m$ .

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2 years ago

The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

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