Answer:
Explanation:
Group one:
The elements of group one shows +1 charge because these all are metals and lose their one valance electron.
Hydrogen lithium sodium potassium rubidium cesium francium
Group 2:
The elements of group two shows +2 charge because these all alkali metals and lose their two valance electrons.
beryllium magnesium calcium strontium barium radium
Group 3:
The elements of group three-B shoes +3 charge by losing three valance electrons.
Scandium yttrium lanthanum actinium
Group 4:
The elements of group 4th A and 4th B lose four electrons or gain four electrons to complete the octet and shows +4 or -4 charge.
Group 5:
Group 5th elements gain three electrons and shows -3 charge to complete the 8 electrons. (octet).
It involve the elements of group 5th A.
Group 6:
The elements of group 6A gain two electrons to complete the octet and shows -2 charge.
Group 7:
The elements of group 7A gain one electron to complete the octet and shows -1 charge.
Group 8:
The elements of group 8A are noble gases and have complete octet. That's why shows 0 charge.
The atomic number of Li is 3
Electron configuration of Li : 1s² 2s¹
The atomic number of Na is 11
Electron configuration of Na : 1s²2s²2p⁶3s¹
Thus there is one electron in the valence shell of Li (2s¹) and that of Na (3s¹). However, the valence electron in Na is in a shell that is farther away from the nucleus compared to that of Li. As a result, the Na valence electron will be held less tightly by the nucleus i.e. it will experience a reduced nuclear attraction and can be removed easily than the Li 2s electron.
8.8 × 10-5 M is the [H3O+] concentration in 0.265 M HClO solution.
Explanation:
HClO is a weak acid and does not completely dissociate in water as ions.
the equation of dissociation can be written and ice table to be formed.
HClO +H2O ⇒ ClO- + H3O+
I 0.265 0 0
C -x +x +x
E 0.265-x +x +x
Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.
Ka = ![\frac{[ClO-][H3O+]}{[HClO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BClO-%5D%5BH3O%2B%5D%7D%7B%5BHClO%5D%7D)
2.9 × 10^-8 = ![\frac{[x] [x]}{[0.265-x]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bx%5D%20%5Bx%5D%7D%7B%5B0.265-x%5D%7D)
= 7.698 x
x = 8.8 × 10-5 M
The hydronium ion concentration is 8.8 × 10-5 M in 0.265 M solution of HClO.
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.
Answer:
28.2
Explanation:
Add all of the pressures, 55, 90, and 50, and divide 100 by the answer you get (195). You'll get 0.512820513 and multiply it by .55 (atm of Oxygen) and you'll get 28.2
