<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>
The arrangement of particles in a gas is random. they have no orderly arrangement and are free to move around while the particles in solid are in an orderly and rigid arrangement and cannot move about. particles in liquid are also arranged orderly but are not rigid
What is the percent composition by mass of oxygen in magnesium oxide, MgO?
Answer: 39.7 percent
Calculate the pOH first :
pOH =14-8.57=5.43
(OH)=10^-5.47
(OH)=3.72 x 10^-6 mol/L
