The Balanced chemical equation of reaction of Borane with oxygen is as follow,
B₂H₆ + 3O₂ -----> 2HBO₂ + 2H₂O
According to this equation 27.66 g (1 mole) of B₂H₆ reacts with oxygen to produce 36 g (2 moles) of water.
The amount of water produced when 19.2 g of B₂H₆ reacted is calculated as follow,

=

Solving for x,
x = (36 g of H₂O ₓ 19.2 g of H₂B₆) / 27.66 g of B₂H₆
x =
24.98 g of H₂O
Result:
24.98 g of water is produced when 19.2 g of B₂H₆ is reacted with excess of oxygen.
Answer:
I believe the molecular formula is MnBr2·4H2O or Br2H8MnO4
Explanation:
Answer:
High concentration of glutathione should be included in the elution buffer for the given experiment to remove protein from the column.
Explanation:
Affinity chromatography is one of important biochemical technique of chromatography which depends on the affinity of ligand for the receptor.Here the ligand is a protein mixture which act as mobile phase and the receptor is present in the wall of chromatography column act as stationary phase.When the protein mixture is applied on the top of the column the substances present within the protein that have high affinity for the receptor present in the walls of chromatography column binds to the later but rest of the protein pass away through the column.High concentration of ligand is used within an buffer solution to remove the desired protein from the column.
From this point of view it can be stated that in the given question high concentration of glutathione should be used to remove the desired protein from the column.
Answer:
eight
Explanation:
Each shell can contain only a fixed number of electrons: The first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on.
To get the value of ΔG we need to get first the value of ΔG°:
when ΔG° = - R*T*㏑K
when R is constant in KJ = 0.00831 KJ
T is the temperature in Kelvin = 25+273 = 298 K
and K is the equilibrium constant = 4.5 x 10^-4
so by substitution:
∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4
= -19 KJ
then, we can now get the value of ΔG when:
ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]
when ΔG° = -19 KJ
and R is constant in KJ = 0.00831
and T is the temperature in Kelvin = 298 K
and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m
so, by substitution:
ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )
= -40