POH of a 0.0072 M=-lg(0.0072) = 2.1426675
the answer is in the picture
Answer:
the mole fraction of Gas B is xB= 0.612 (61.2%)
Explanation:
Assuming ideal gas behaviour of A and B, then
pA*V=nA*R*T
pB*V=nB*R*T
where
V= volume = 10 L
T= temperature= 25°C= 298 K
pA and pB= partial pressures of A and B respectively = 5 atm and 7.89 atm
R= ideal gas constant = 0.082 atm*L/(mol*K)
therefore
nA= (pA*V)/(R*T) = 5 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 2.04 mole
nB= (pB*V)/(R*T) = 7.89 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 3.22 mole
therefore the total number of moles is
n = nA +nB= 2.04 mole + 3.22 mole = 5.26 mole
the mole fraction of Gas B is then
xB= nB/n= 3.22 mole/5.26 mole = 0.612
xB= 0.612
Note
another way to obtain it is through Dalton's law
P=pB*xB , P = pA+pB → xB = pB/(pA+pB) = 7.69 atm/( 5 atm + 7.89 atm) = 0.612
Answer:

Explanation:
Hello!
In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Which can be also written as:

Thus, since we need the specific heat of the metal, we solve for it as shown below:

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If one or more of the following occur: Gas formation, precipitate formation, color change, temperature change, or odor