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:
NH₃/NH₄Cl
Explanation:
We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.
![pH=pKa+log\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.
Let's consider the following buffers and their pKa.
- CH₃COONa/CH3COOH (pKa = 4.74)
The optimum buffer is NH₃/NH₄Cl.
Answer:
= 459.1 nm
This wavelength corresponds to yellow color and thus gold has warm yellow color.
Explanation:
Given that:- Energy = 2.7 eV
Energy in eV can be converted to energy in J as:
1 eV = 1.602 × 10⁻¹⁹ J
So, Energy = 
Considering:-
Where,
h is Plank's constant having value
c is the speed of light having value
is the wavelength of the light
So,

= 459.1 nm
This wavelength corresponds to yellow color and thus gold has warm yellow color.
After some thinking I have come to the conclusion that the answer is C.