Answer:
1.59 x 10⁻²⁵ J.
Explanation:
- The energy of a photon is calculated Planck - Einstein's equation:
E = h ν
, where
E is the energy of the photon,
h is Planck's constant <em>(h = 6.626 x 10
⁻³⁴ J.s)</em>
ν is the frequency of the photon
-
There is a relation between the frequency (ν
) and wave length (λ).
λ.ν = c,
where c is the speed of light in vacuum (c = 3
.0 x 10
⁸ m/s).
λ = 125 cm = 1.25 m.
<em>Now, E = h.c/λ.</em>
∴ E = h.c/λ = (6.626 x 10
⁻³⁴ J.s) (3
.0 x 10
⁸ m/s) / (1.25 m) = 1.59 x 10⁻²⁵ J.
Answer:
Problem Details
The element europium exists in nature as two isotopes: 151Eu has a mass of 150.9196 amu, and 153Eu has a mass of 152.9209 amu. The average atomic mass of europium is 151.96 amu. Calculate the relative abundance of the two europium isotopes.
answer:
151Eu = 48%, 153Eu = 52%
The substance that can not be broken down by chemical means is antimony.
t can be transferred from one region to the other. In solids heat is transferred by a process called conduction. In vacuum heat transfer is by a process called radiation. In fluids (gases and liquids) heat is transferred by a process known as convection.
This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:
<h3>Heating curves:</h3>
In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.
Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:
![Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\ \\ Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\ \\ Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\ \\ Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\ \\ Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ](https://tex.z-dn.net/?f=Q_1%3D0.0217mol%2A111.5%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%28-114.1%5C%C2%B0C%29-%28-200%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.208kJ%5C%5C%0A%5C%5C%0AQ_2%3D0.0217mol%2A4.9%5Cfrac%7BkJ%7D%7Bmol%7D%20%3D0.106kJ%5C%5C%0A%5C%5C%0AQ_3%3D0.0217mol%2A112.4%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%2878.4%5C%C2%B0C%29-%28-114.1%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.470kJ%5C%5C%0A%5C%5C%0AQ_4%3D0.0217mol%2A38.6%5Cfrac%7BkJ%7D%7Bmol%7D%20%3D0.838kJ%5C%5C%0A%5C%5C%0AQ_5%3D0.0217mol%2A87.5%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%28150%5C%C2%B0C%29-%2878.4%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.136kJ)
Finally, we add them up to get the result:

Learn more about heating curves: brainly.com/question/10481356