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3 years ago

Calculate the energy of light having a wavelength of 395 nm. show work

Chemistry

1 answer:

Liono4ka [1.6K]3 years ago

6 0

Answer: $5.032\times 10^{-19}\ J$

Explanation:

Given

The wavelength of the light is $\lambda =395\ nm$

Energy associated with a photon(light) is given by $E=\frac{hc}{\lambda}$

where, $h=\text{Planck constant}$

$c=\text{speed of light}$

$\Rightarrow E=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{395\times 10^{-9}}\\\\\Rightarrow E=5.032\times 10^{-19}\ J$

Therefore, the energy associated with the light is $5.032\times 10^{-19}\ J$

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Answer:

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$b=\frac{n_{solute}}{m_{solvent}(kg)}$

If you place 5 moles of sodium chloride and 4 moles of sucrose into 11 moles of water, the mole fraction of sodium chloride would be 0.25

$X(NaCl)=\frac{molesNaCl}{moles_{solution}} =\frac{5mol}{20mol} =0.25$

A solution is made up of 0.15 grams of sodium chloride in 1 liter of water. For this solution, the solvent is water.

The solvent is the component in major proportion and the one that defines the state of the solution.

A solution is made up of 0.15 grams of sodium chloride in 1 liter of water. For this solution, the solute is sodium chloride.

The solute is the component in minor proportion.

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3 years ago

A lab scientist cools a liquid sample of water (2.6 kg) at 0.00°C to -192°C. The water turns to ice as this temperature change o

soldier1979 [14.2K]

Answer: The heat released for the given process is -1892 kJ

Explanation:

The processes involved in the given problem are:

$1.)H_2O(l)(0^oC,273K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(s)(-192^oC,81K)$

Pressure is taken as constant.

To calculate the amount of heat released at same temperature, we use the equation:

$q=m\times L_{f,v}$ ......(1)

where,

q = amount of heat released = ?

m = mass of water/ice

$L_{f,v}$ = latent heat of fusion or vaporization

To calculate the amount of heat released at different temperature, we use the equation:

$q=m\times C_{p,m}\times (T_{2}-T_{1})$ .......(1)

where,

q = amount of heat released = ?

$C_{p,m}$ = specific heat capacity of medium

m = mass of water/ice

$T_2$ = final temperature

$T_1$ = initial temperature

Calculating the heat absorbed for each process:

For process 1:

Converting the latent heat of fusion in J/kg, we use the conversion factor:

1 kJ = 1000 J

So, $(\frac{-334kJ}{1kg})\times (\frac{1000J}{1kJ})=-334\times 10^3J/kg$

We are given:

$m=2.6kg\\L_f=-334\times 10^3J/kg$

Putting values in equation 1, we get:

$q_1=2.6kg\times (-334\times 10^3J/kg)=-868400J$

For process 2:

We are given:

$m=2.6kg\\C_{p,s}=2050J/kg.K\\T_1=273K\T_2=81K$

Putting values in equation 2, we get:

$q_2=2.6kg\times 2050J/kg.K\times (81-(273))^oC\\\\q_2=1023360J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[-868400+(-1023360)]J=-1891760J=-1891.76kJ\approx -1892kJ$

Hence, the heat released for the given process is -1892 kJ

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3 years ago