Answer is: there is 2,69·10²³ atoms of bromine.
m(CH₂Br₂) = 39,0 g.
n(CH₂Br₂) = m(CH₂Br₂) ÷ M(CH₂Br₂).
n(CH₂Br₂) = 39 g ÷ 173,83 g/mol.
n(CH₂Br₂) = 0,224 mol.
In one molecule of CH₂Br₂, there is two bromine atoms, so:
n(CH₂Br₂) : n(Br) = 1 : 2.
n(Br) = 0,448 mol.
N(Br) = n(Br) · Na.
N(Br) = 0,448 mol · 6,022·10²³ 1/mol.
n(Br) = 2,69·10²³.
Answer:
162.7miles/hr
Explanation:
Given parameters:
Distance covered by the train = 813.5miles
Time taken = 5hours
Unknown:
Speed of the train = ?
Solution:
Speed is a physical quantity.
It is mathematically expressed as;
Speed = 
So, input parameters and solve;
Speed = 
= 162.7miles/hr
Because it’s kinetic energy INCREASES the speed
Answer:
A.) ![K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
Explanation:
The general Kb expression is:
![K_b = \frac{[HA][OH^-]}{[A^-]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BHA%5D%5BOH%5E-%5D%7D%7B%5BA%5E-%5D%7D)
In this equation
-----> Kb = equilibrium constant
-----> [HA] = acid
-----> [A⁻] = base
Since liquids are not included in equilibrium expressions, H₂O should not be present. The products are in the numerator while the reactant are in the denominator. In this reaction, CH₃NH₂ is acting as a base and CH₃NH₃⁺ is acting as an acid.
As such, the expression is:
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
