Answer:
Stone A
Explanation:
Measuring density is an easiest way to determine if two similar looking substances are same or not. Here also we need to perform the density test for each stone that is suspected to be fake diamond. We will calculate the density of each stone and compare it with the density of original diamond.
Density is calculated using the formula
It has been given in the question that both the substances have same volume of 0.15 
.
Density of stone A = 
= 3.47 
or after rounding off we get 3.5
Density of stone B = 
= 2.8
It is clear from the above calculation that the stone A has same density as the diamond but stone B lacks behind in density.
So, stone A could be the real diamond.
IT will be Refraction of light
Answer:
Equilibrium constant for the reaction is 0.0892
Explanation:
This is the reaction of equilibrium
H₂O + Cl₂O ⇄ 2HClO
Eq 0.077M 0.077M 0.023M
Let's make the expression for Kc
Kc = [HClO]² / [Cl₂O] . [H₂O]
Kc= 0.023² / 0.077 . 0.077 = 0.0892
<u>Answer:</u> The increase in pressure is 0.003 atm
<u>Explanation:</u>
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= final pressure = ?
= Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![801^oC=[801+273]K=1074K](https://tex.z-dn.net/?f=801%5EoC%3D%5B801%2B273%5DK%3D1074K)
= final temperature = ![(801+1.00)^oC=802.00=[802+273]K=1075K](https://tex.z-dn.net/?f=%28801%2B1.00%29%5EoC%3D802.00%3D%5B802%2B273%5DK%3D1075K)
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28800J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B1074%7D-%5Cfrac%7B1%7D%7B1075%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D3%5Ctimes%2010%5E%7B-3%7Datm%5C%5C%5C%5CP_2%3De%5E%7B3%5Ctimes%2010%5E%7B-3%7D%7D%3D1.003atm)
Change in pressure = 
Hence, the increase in pressure is 0.003 atm
Hydrocarbon hydrogen thats all ik sorry
