A tube of air contains a feather and a coin. both objects are dropped at the same time. the coin falls faster than the feather.
1 answer:
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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
Answer:
Explanation:
We are given the percent composition: 22.5% phosphorus and 77.5% chlorine.
We can assume there are 100 grams of this compound. We choose 100 because we can simply use the percentages as the masses.
- 22.5 g P
- 77.5 g Cl
Next, convert these masses to moles, using the molar masses found on the Periodic Table.
- P: 30.974 g/mol
- Cl: 35.45 g/mol
Use the molar masses as ratios and multiply by the number of grams.
Divide both of the moles by the smallest number of moles to find the mole ratio.
The mole ratio is about 1 P: 3 Cl, so the empirical formula is written as:<u> PCl₃</u>