Lava I think try that hope it helps :)
Answer:
a. Remaining at rest requires the use of ATP.
Explanation:
The resting membrane potential is maintained by the sodium-potassium pump. The sodium potassium pump does this by actively pumping sodium ions out of the cell and potassium ions inside the cell in a ratio of 3:2. This movement of ions by the sodium-potassium pump is against their concentration gradient. In a neuron at rest, there are more sodium ions outside the cell than there are inside the cell. Also, there are are more potassium ions inside the cell than there are outside the cell. However, there are ion channels through which these ions enter and leave the cell. Sodium ion channels allow sodium to enter the cell following its concentration gradient, whereas, potassium ion channels allow potassium to leave the cell following its concentration gradient. However, more potassium ions leave the cell than do sodium ions enter the cell because of the higher permeability of the cell to potassium ions.
In order to maintain the resting membrane potential, the sodium potassium pump powered by the hydrolysis of an ATP molecules pumps sodium ions out of the cell and potassium ions into the cell.
<em>Therefore, the correct option is A, as ATP is needed by the sodium-potassium pump in order to maintain the resting membrane potential.</em>
I think the most appropriate answer is: the solvent being used in the experiment
<span>To correct for any light absorption not originating from the solute you will need to calibrate the tools with a solution that most similar to the sample.
Blank covete or standard solution can be used, but it was not ideal. By using the solvent as calibration, you can remove the reading from the solvent so your result only comes from the sample.
</span>
The molarity of potassium ions in a 0.122 M K2CrO4 is 0.244 M
<u><em>Explanation</em></u>
write dissociation reaction for K2CrO4
that K2CrO4 (aq)→ 2K^+ (aq) + CrO4^2- (aq)
K2CrO4 dissociate to give 2 ions of potassium ,therefore the molarity of potassium ion = 2 x 0.122 = 0.244 M
