Answer:
Ka= [H+][But-]/H[Hbut]
Let k moles of H+ be formed
1.52*10^-5=k^2/0.004-k
K^2=0.004*1.52*10^-5-1.52*10^-5k
K^2- 1.52*10^-5k-6.08*10^-8=0
Solving this quadratic equation.
K=7.6*10^-6
PH=-log(7.6*10^-6)=5.2
2.04 atm is the pressure of the nitrogen gas in the container when 0.316 mol sample of nitrogen gas, N2 (g), is placed in a 4.00 L container at 315 K.
Explanation:
Data given:
moles of nitrogen gas, n = 0.316
volume of the nitrogen gas, V = 4 litres
temperature of the container of nitrogen gas = 315 K
R (gas constant) = 0.08201 Latm/mole K
Pressure of the nitrogen gas on the container, P = ?
from the data given, we will apply ideal gas law equation to calculate the pressure on the nitrogen gas:
PV = nRT
Rearranging the equation,
P = 
P = 
P = 2.04 atm
The pressure of the nitrogen gas in the container is 2.04 atm.
Answer:
A. 444.5 pm
Explanation:
We know that:

i.e.


in a face-centered cubic crystal, the number of atoms per unit cell is (n) = 4
The molar mass of manganese (II) oxide ![[Mn(11)O] = 70.93 \ g/mol](https://tex.z-dn.net/?f=%5BMn%2811%29O%5D%20%3D%2070.93%20%5C%20g%2Fmol)
Density
is given as 5.365 g/cm³
Avogadro constant
= 6.023 × 10²³ atoms/mol
∴

Making th edge length "a" the subject, we get:



![a= \sqrt[3]{8.78 \times 10^{-23} \ cm^3}](https://tex.z-dn.net/?f=a%3D%20%5Csqrt%5B3%5D%7B8.78%20%5Ctimes%2010%5E%7B-23%7D%20%5C%20cm%5E3%7D)
a = 4.445 × 10⁻⁸ cm
a = 444.5 pm
Answer:
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Explanation:
I don’t think they have to
Explanation:
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