Explanation:
The pure form of water has a boiling point of 100°C. Boiling point is a physical property of matter and it shows that at such temperature, a liquid substance will change state to vapor.
Pure water is made up of 2 atoms of hydrogen and 1 atom of oxygen. The only intermolecular forces between them is the hydrogen bonds that must be broken for the water to boiling off.
In hydrate, water is present in another form. The water is attached to another compound.
For a pure liquid, the they have reasonably constant boiling point and low boiling range.
Impurities such as the other part of the hydrate causes the elevation of the boiling point and the widening of the boiling range for impure substances.
We are no longer dealing with just hydrogen bonds, other molecular interactions are now involved and they need to be accounted for.
learn more:
Pure substances brainly.com/question/1832352
#learnwithBrainly
Isotopes of the same element
₃₅⁷⁷X and ₃₅⁸¹X
<h3>Further explanation</h3>
Given
Isotopes of element
Required
Isotopes of the same element
Solution
The elements in nature have several types of isotopes
Isotopes are elements that have the same Atomic Number (Proton) and different mass numbers
Element symbols that meet the 2 conditions above are :
₃₅⁷⁷X and ₃₅⁸¹X
This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and
and
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Finally we convert this result to kJ:

Learn more:
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
Answer:
science ka Hindi kya hoga