Using the periodic table, choose the more reactive non-metal. S or As
1 answer:
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Answer:
The correct answer is option B.
Explanation:
Moles of = 40 mol
Moles of NaOH = 48 mol
According to reaction, 3 moles of NaOH reacts with 2 moles
Then ,48 moles of NaOH will reacts with:
of
Then ,40 moles of will reacts with:
of NaOH
As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.
Moles left after reaction = 40 mol - 32 mol = 8 mol
Hence, the is an excessive reagent.
Answer:
The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.
Explanation:
The free change for the process can be written in terms of its equilibrium constant as:
ΔG° =
where:
R= universal gas constant
T= temperature
= equilibrum constant for the process
Similarly, free energy change and cell potentia; are related to each other as follows;
ΔG= -nFE°
from above;
F = faraday's constant
n = number of electrons exchanged in the process; and
E = standard cell potential
∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:
ΔG° =
where;
[texK_eq[/tex]=
For transport of ions to an internal pH of 7.4, the transport taking place can be given as:
⇒
Equilibrum constant for the transport is given as:
= 10⁻⁷⁴
=3.98 * 10⁻⁸M
= 10⁻²¹
=7.94 * 10⁻³M
Hence;
=
= 5.012 × 10⁻⁶
Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:
ΔG° =
If temperature T= 37° C ; in kelvin
=37° C + 273.15K
=310.15K; and
R-= 8.314 j/mol/k
substituting the values into the equation we have;
ΔG₁ =
= 31467.93Jmol⁻¹
≅ 31.47KJmol⁻¹
If the potential difference across the cell membrane= 60.0mV.
Energy required to cross the cell membrane will be:
ΔG₂ =
ΔG₂ =
= 5.79KJ
Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹
Now, we can calculate the total amount of energyy required to transport H⁺ ions across the membrane:
Δ
= (31.47+5.79) KJmol⁻¹
= 37.26KJmol⁻¹
We can therefore conclude that;
The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹