Metal usually donates electrons. The concept behind this phenomenon is stability. The elements with the most stable electronic configuration are the noble gases in Group 5A. As a result, the other elements donate or accept electrons so that they would be like the noble gases. Since metals are past their nearest noble gas element, they have to shed their electrons. When they do, they become cations which are positively charged ions.
Answer:
0.203 is the mean of the concentration of the HCl solution
Explanation:
You have 5 concentrations. The most appropiate result is the mean of those results. The mean is a statistical defined as the sum of each result divided by the total amount of results. For the results of the problem, the mean is:
0.210 + 0.204 + 0.201 + 0.202 + 0.197 = 1.014 / 5 =
<h3>0.203 is the mean of the concentration of the HCl solution</h3>
Answer:
3.5 atm
Explanation:
As stated in the question pressure is required to counteract the natural tendency for water to dilute the more concentrated solution. The difference in concentrations will give us the answer using the osmotic pressure equation.
π = ( n/v) RT where n/v is the molarity (mol/L), R is the gas constant and T is the temperature.
The difference in osmotic pressure of the solutions is:
Δπ = Δ c RT where c is the difference in molar concentrations.
pressure required = Δπ = (0.190 - 0.048) M x 0.821 Latm/Kmol x 298 K
= 3.47 atm
Answer:
The correct answer is "Secondary active transport".
Explanation:
Secondary active transport is a form of across the membrane transport that involves a transporter protein catalyzing the movement of an ion down its electrochemical gradient to allow the movement of another molecule or ion uphill to its concentration/electrochemical gradient. In this example, the transporter protein (antiporter), move 3 Na⁺ into the cell in exchange for one Ca⁺⁺ leaving the cell. The 3 Na⁺ are the ions moved down its electrochemical gradient and the one Ca⁺⁺ is the ion moved uphill its electrochemical gradient, because Na+ and Ca⁺⁺are more concentrated in the solution than inside the cell. Therefore, this scenario is an example of secondary active transport.
6.28×1013+7.30×1011 this =13741.94
