Answer:
ON
Explanation:
Frictional force is a force that opposes the motion of body. It is because of friction that motion can be controlled and even walking is made possible.
Now, on a frictionless surface, there is no opposition to motion. Therefore, to keep the body moving there is no force required.
- If surface lacks friction, motion will continue perpetually without any hinderance.
First, we will use the general gas formula to get the number of moles.
PV = nRT where:
P is the pressure of gas = 751 mmHg = 100125.096375 Pascal
V is the volume = 1 liter = 0.001 m^3
n is the number of moles we want to calculate
R is the gas constant = <span>8.314 J/(K. </span>mol<span>)
T is the temperature = 31 degrees celcius = </span>304.15 degree kelvin
Substitute in the above equation to get the number of moles as follows:
100125.096375 * 0.001 = n * 8.314 * 304.15
n = 0.039595 moles
Now, we will use the number of moles to get the mass as follows:
number of moles = mass / molar mass
mass = number of moles * molar mass
number of moles = 0.039595 moles
molar mass of ammonia (NH3) = 14 + 3(1) = 17 grams
Substitute to get the mass as follows:
mass = 0.039595 * 17 = 0.673122 grams
Last step is to get the density as follows:
density = mass / volume
mass = 0.673122 grams
volume = 1 liter
density = 0.673122 / 1 = 0.673122 grams/liter = <span>0.000675 kg/L</span>
The dissociation constant of the base : 7.4 x 10⁻⁴
<h3>Further explanation</h3>
Butylamine, C4H9NH2 Is A Weak Base
Kb is the dissociation constant of the base.
LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)
![\rm Kb=\dfrac{[L][OH^-]}{[LOH]}](https://tex.z-dn.net/?f=%5Crm%20Kb%3D%5Cdfrac%7B%5BL%5D%5BOH%5E-%5D%7D%7B%5BLOH%5D%7D)
[OH⁻] for weak base can be formulated :
![\tt [OH^-]=\sqrt{Kb.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D)
pH of solution : 12
pH+pOH=14, so pOH :
14-12 = 2, then :
![\tt [OH^-]=10^{-pOH}\\\\(OH^-]=10^{-2}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D10%5E%7B-pOH%7D%5C%5C%5C%5C%28OH%5E-%5D%3D10%5E%7B-2%7D)
the the dissociation constant (Kb) =
Or you can use from ICE method :
C4H9NH2(aq) + H2O(l) ⇌ C4H9NH3+(aq) + OH-(aq)
0.15
x x x
0.15-x x x
![\tt Kb=\dfrac{x^2}{0.15-x}\rightarrow x=[OH^-]\\\\Kb=\dfrac{10^{-4}}{0.15-10^{-2}}=7.14\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20Kb%3D%5Cdfrac%7Bx%5E2%7D%7B0.15-x%7D%5Crightarrow%20x%3D%5BOH%5E-%5D%5C%5C%5C%5CKb%3D%5Cdfrac%7B10%5E%7B-4%7D%7D%7B0.15-10%5E%7B-2%7D%7D%3D7.14%5Ctimes%2010%5E%7B-4%7D)
<span>the hydrophobic portions would surround the oil and hydrophilic would surround the water... for either</span>
Answer:
The reaction type is Synthesis
