Maybe 4mL cause it sounds like a math problem so that should be it
I believe the answer to this question is true
Answer:
The value ![N_A = 0.192 \ mol \cdot m^{-2} \cdot \ s](https://tex.z-dn.net/?f=N_A%20%3D%200.192%20%5C%20mol%20%5Ccdot%20m%5E%7B-2%7D%20%5Ccdot%20%5C%20%20s)
Explanation:
From the question we are told that
The thickness of the air is ![z_2 - z_1 = 1 \ cm =0.01 \ m](https://tex.z-dn.net/?f=z_2%20-%20z_1%20%20%3D%20%201%20%5C%20cm%20%20%3D0.01%20%5C%20%20m)
The temperature is ![T = 25^oc = 25 +273 = 298 \ K](https://tex.z-dn.net/?f=T%20%3D%2025%5Eoc%20%3D%2025%20%2B273%20%3D%20298%20%5C%20%20K)
The total pressure is ![P_T = 1 atm = 1.01325*10^{5} \ Pa](https://tex.z-dn.net/?f=P_T%20%3D%201%20atm%20%20%3D%201.01325%2A10%5E%7B5%7D%20%5C%20%20Pa)
The partial pressure of Ammonia first side is
The partial pressure of Ammonia to the second side is ![P_{A} = 0.1 \ atm = 0.1 * 1.0325*10^{5} = 10132.5 \ Pa](https://tex.z-dn.net/?f=P_%7BA%7D%20%3D%200.1%20%5C%20atm%20%20%3D%200.1%20%2A%201.0325%2A10%5E%7B5%7D%20%3D%2010132.5%20%5C%20Pa)
Rate of flow of ammonia is
![D_{AB} = 0.214 \ cm/s = \frac{0.214 }{10000} = 2.14 *10^{-5} \ m^2 /s](https://tex.z-dn.net/?f=D_%7BAB%7D%20%3D%200.214%20%5C%20cm%2Fs%20%3D%20%5Cfrac%7B0.214%20%7D%7B10000%7D%20%3D%202.14%20%2A10%5E%7B-5%7D%20%5C%20%20m%5E2%20%2Fs)
Generally the molar flux of ammonia is mathematically represented as
![N_A = \frac{D_{AB} * P_T }{RT(z_2 -z_1)} * ln [\frac{P_T - P_{Al}}{P_T - P_{AO}} ]](https://tex.z-dn.net/?f=N_A%20%3D%20%5Cfrac%7BD_%7BAB%7D%20%2A%20P_T%20%7D%7BRT%28z_2%20-z_1%29%7D%20%2A%20ln%20%5B%5Cfrac%7BP_T%20-%20P_%7BAl%7D%7D%7BP_T%20-%20P_%7BAO%7D%7D%20%5D)
Here R is the gas constant with value
![R = 8.314 \ m^3 \cdot Pa \cdot mol^{-1} \cdot K](https://tex.z-dn.net/?f=R%20%3D%208.314%20%5C%20m%5E3%20%5Ccdot%20Pa%20%5Ccdot%20mol%5E%7B-1%7D%20%5Ccdot%20K)
![N_A = \frac{2.14 *10^{-5} * 1.01325*10^{5} }{8.314 *298 (0.01)} * ln [\frac{1 - 0.1}{1 - 0.9} ]](https://tex.z-dn.net/?f=N_A%20%3D%20%5Cfrac%7B2.14%20%2A10%5E%7B-5%7D%20%20%2A%201.01325%2A10%5E%7B5%7D%20%20%7D%7B8.314%20%2A298%20%280.01%29%7D%20%2A%20ln%20%5B%5Cfrac%7B1%20%20-%200.1%7D%7B1%20-%200.9%7D%20%5D)
=> ![N_A = 0.192 \ mol \cdot m^{-2} \cdot \ s](https://tex.z-dn.net/?f=N_A%20%3D%200.192%20%5C%20mol%20%5Ccdot%20m%5E%7B-2%7D%20%5Ccdot%20%5C%20%20s)
Answer:
E = 1053.365 J
Explanation:
∴ Q = 440μC * ( 1 E-6 C / μC ) = 4.4 E-4 C
∴ C = k*εo*A / d
∴ k mica = 5.4
∴ εo = 8.8542 E-12 C² / N.m²
∴ A = 6.2cm * 6.2 cm = 38.44 cm² * ( m/100cm)² = 3.844 E-3 m²
∴ d = 2mm * ( m / 1000mm ) = 2 E-3 m
⇒ C = (( 5.4 ) * (8.8542 E-12 ) * ( 3.844 E-3 )) / 2 E-3
⇒ C = 9.1896 E-11 C²/N.m
⇒ E = ( 4.4 E-4 )² / (2*9.1896 E-11)
⇒ E = 1053.365 N.m = 1053.365 J
Data Given:
[H⁺] = 3.16 × 10⁻⁷ M
To Find:
pH = ?
Solution:
As we know that,
pH = -log [H⁺]
Putting values,
pH = -log (3.16 × 10⁻⁷)
pH = 6.50
Kind of Solution:
As we know, when pH of a solution is equal to 7, it is Neutral. When less than 7, its Acidic and when greater than 7, it is Basic. As this solution has pH 6.50, which is less than 7, so it is slightly Acidic in Nature.