Answer:
-30.7 kj/mol
Explanation:
The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ’= -RTln K’eq
where,
R = -8.315 J / mo
T = 298 K
For reaction,
1. K′eq1=270,
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 270
= - 8.315 x 298 x 5.59
= - 13,851.293 J / mo
= - 13.85 kj/mol
2. K′eq2=890
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 890
= - 8.315 x 298 x 6.79
= - 16.82 kj/mol
therefore, total standard free energy
= - 13.85 + (-16.82)
= -30.7 kj/mol
Thus, -30.7 kj/mol is the correct answer.
There would be eighteen bonding electrons
Answer:
Se detailed explanation.
Explanation:
Hello,
In this case, since both magnesium and calcium ions are in group IIA, we can review the following similar properties:
- Since both calcium and magnesium are in group IIA they have two valence electrons, it means that the both of them have two electrons at their outer shells.
- They are highly soluble in water when forming ionic bonds with nonmetals such as those belonging to halogens and oxygen's family.
- Calcium has 18 electrons and magnesium 10 which are two less than the total protons (20 and 12 respectively) since the both of them have lost two electrons due their ionized form.
- Their electron configurations are:

It means that the both of them are at the
region since it is the last subshell at which their electrons are.
Best regards.
Answer:I dont know the answer but i need the points thx!!!!
Explanation:
Answer:
The ideal gas law is expressed mathematically by the ideal gas equation as follows;
P·V = n·R·T
Where;
P = The gas pressure
V = The volume of the gas
n = The number of moles of the gas present
R = The universal gas constant
T = The temperature of the gas
A situation where the ideal gas law is exhibited is in the atmosphere just before rainfall
The atmospheric temperature of the area expecting rainfall drops, (when there is appreciable blockage of the Sun's rays by cloud covering) followed by increased wind towards the area, which indicates that the area was in a state of a low pressure, 'P', and or volume, 'V', or a combination of both low pressure and volume P·V
When the entry flow of air into the area is observed to have reduced, the temperature of the air in the area is simultaneously sensed to have risen slightly, therefore, the combination of P·V is seen to be proportional to the temperature, 'T', and the number of moles of air particles, 'n' in the area
Explanation: