<span>We can use the ideal gas law PV=nRT
For the first phase
The starting temperature (T1) is 273.15K (0C). n is 1 mole, R is a constant, P = 1 atm, V1 is unknown.
The end temperature (T2) is unknown, n= 1 mol, R is a constant, P = 3*P1= 3 atm, V2=V1
Since n, R, and V will be constant between the two conditions: P1/T1=P2/T2
or T2= (P2*T1)/(P1) so T2= (3 atm*273.15K)/(1 atm)= 3*273.15= 816.45K
For the second phase:
Only the temperature and volume change while n, P, and R are constant between the start and finish.
So: V1/T1=V2/T2 While we don't know the initial volume, we know that V2=2*V1 and T1=816.45K
So T2=(V2*T1)/V1= (2*V1*T1)/V1=2*T1= 2*816.45K= 1638.9K
To find the total heat added to the gas you need to subtract the original amount of heat so
1638.9K-273.15K= 1365.75K</span>
From ideal gas equation that is PV=nRT
n(number of moles)=PV/RT
P=760 torr
V=4.50L
R(gas constant =62.363667torr/l/mol
T=273 +273=298k
n is therefore (760torr x4.50L) /62.36367 torr/L/mol x298k =0.184moles
the molar mass of NO2 is 46 therefore density= 0.184 x 46=8.464g/l
Suspension. The particles are big enough for the eye to see, and will separate if left sitting.
Answer:
Sure!
Explanation:
I could use someone to talk to so I don't see why not.
<h3>Answer:</h3>
Rubidium (Rb)
<h3>Explanation:</h3>
Ionization Energy is defined as, "the minimum energy required to knock out or remove the valence electron from valence shell of an atom".
<h3>Trends in Periodic table:</h3>
Along Periods:
Ionization Energy increases from left to right along the periods because moving from left to right in the same period the number of protons (atomic number) increases but the number of shells remain constant hence, resulting in strong nuclear interactions and electrons are more attracted to nucleus hence, requires more energy to knock them out.
Along Groups:
Ionization energy decreases from top to bottom along the groups because the number of shells increases and the distance between nucleus and valence electrons also increases along with increase in shielding effect provided by core electrons. Therefore, the valence electrons experience less nuclear attraction and are easily removed.
<h3>Conclusion:</h3>
Given elements belong to same group hence, Rubidium present at the bottom of remaining elements will have least ionization energy due to facts explained in trends of groups above.
