Answer:
They have the same number of electron energy levels.
They transition from a metal to noble gas.
Explanation:
Periods in the periodic table of elements refer to elements in the same row. All the elements in a certain row of the periodic table;
have the same number of electron energy levels.
transition from a metal to noble gas.
Answer:
16
Explanation:
Group two elements are alkaline earth metal.
All these have two valance electrons. In order to achieve noble gas configuration it loses its two valance and get complete octet.
Reaction with group 16.
Reaction with oxygen,
They react with oxygen and form oxide.
2Ba + O₂ → 2BaO
2Mg + O₂ → 2MgO
2Ca + O₂ → 2CaO
this oxide form hydroxide when react with water,
BaO + H₂O → Ba(OH)₂
MgO + H₂O → Mg(OH)₂
CaO + H₂O → Ca(OH)₂
With sulfur,
Mg + S → MgS
Ca + S → CaS
Ba + S → BaS
We know that organisms inherit their traits from their parents, and these traits are a combination of the traits their parents possessed. Therefore, by using a pedigree to map the ancestry of an organism, we may evaluate the propagation of a specific trait through the organism's family. An example of this is when people are assessed for the risk of diseases like breast cancer and sickle cell anemia.
<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
