Answer:
One ATP molecule's hydrolysis can move 3 ions of Sodium (Na+) across the menbrane.
Explanation:
As you can see, the energy provided by ATP is enough for moving 3 ions of Na+. Each ion needs +2.1 kcal/mol of energy.
If we multiply by 3 the energy for moving across the membrane= +6.3 kcal/mol
By adding the energy from ATP:
ΔGTotal=6.3-7.3= -1 kcal/mol
Answer:
Two non bonded electron pairs and four bonded electron pairs
Explanation:
An image of the compound as obtained from chemlibretext is attached to this answer.
The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.
Answer:
Ethynylcyclopropane is the stable isomer for given alkyne.
Explanation:
In order to solve this problem we will first calculate the number of Hydrogen atoms. The general formula for alkynes is as,
CₙH₂ₙ₋₂
Putting value on n = 5,
C₅H₂.₅₋₂
C₅H₈
Also, the statement states that the compound contains one ring therefore, we will subtract 2 hydrogen atoms from the above formula i.e.
C₅H₈ ------------(-2 H) ----------> C₅H₆
Hence, the molecular formula for given compound is C₅H₆
Below, 4 different isomers with molecular formula C₅H₆ are attached.
The first compound i.e. ethynylcyclopropane is stable. As we know that alkynes are sp hybridized. The angle between C-C-H in alkynes is 180°. Hence, in this structure it can be seen that the alkyne part is linear and also the cyclopropane part is a well known moiety.
Compounds 3-ethylcycloprop-1-yne, <u>cyclopentyne </u>and 3-methylcyclobut-1-yne are highly unstable. The main reason for the instability is the presence of triple bond in three, five and four membered ring. As the alkynes are linear but the C-C-H bond in these compound is less than 180° which will make them highly unstable.
Explanation:
As the given reaction is as follows.
So, according to the balanced equation, it can be seen that rate of formation of will be twice the rate of disappearance of
.
And, it is known that rate of disappearance of reactant will be negative and rate of formation of products will be positive value.
This means that,
Rate of the reaction = -Rate of disappearance of
=
=
= M/s
Therefore, calculate the rate of formation of as follows.
Rate of formation of =
= M/s
Thus, we can conclude that the rate of formation of is
M/s.