Answer:
C. Lymphocytes
Explanation:
All of the following choices are kinds of white blood cells that have a significant role in the body's immune system.
A. Monocytes
Monocytes defend the body against infection by acting as macrophages. They are capable of eating up foreign bodies that may cause infection.
B. Neutrophils
Neutrophils are the most in number in the body's immune system, especially when there is inflammation. They are distributed to different areas where they can move through circulation along blood vessels. They specifically attack antigens.
C. Lymphocytes
Lymphocytes are further divided into two. These are the B cells and the T cells. The B cells are also of two kinds. One of them is the memory B cells, which can remember a foreign body and create antibodies against it to provide for a long-term resistance in case the body gets exposed to the same infectious agent again.
D. Basophils
As part of the immune system, basophils function for preventing blood clots as well as in mediating allergic reactions.
Answer:
B) P₂O₅
Explanation:
A) MgCl₂
Molar mass of Mg = 24
Molar mass of Cl = 35.5
Molar mass = 24 + 2×35.5 = 95 g/mol
B) P₂O₅
Molar mass of P = 31
Molar mass of O = 16
Molar mass = 31 ×2 + 5×16 = 142 g/mol
C) BaCl₂
Molar mass of Ba = 137.33
Molar mass of Cl = 35.5
Molar mass = 137.33 + 2×35.5 = 208.33 g/mol
D) AlCl₃
Molar mass of Al = 27
Molar mass of Cl = 35.5
Molar mass = 27 + 3×35.5 = 133.5
Answer:
Volume
Explanation:
A cubic centimeters(cm³) is a measure of volume that is equal to a cube with width, length and height are all 1 centimeters. then, multiply length, width and height values together, this will give you the volume of the cube. 1 inch = 2.54 centimeters
Answer:
Here's what I get
Explanation:
The three resonance forms of N₂O are shown in the first diagram below (you can also use horizontal dashes to represent the bonding pairs).
To get the formal charge (FC) on the atoms, cut each bond in half, as in the second diagram.
Each atom gets the electrons on its side of the cut.
Formal charge = valence electrons in isolated atom - electrons on bonded atom
FC = VE - BE
(a) In Structure A
Left-hand N:
VE = 5
BE = 1 lone pair (2)+ 3 bonding electrons = 2 + 3 = 5
FC = 5 - 5 = 0.
Central N:
VE = 5
BE = 4
FC = 5 - 4 = +1
On O:
VE = 6
BE = 3 lone pairs(6) + 1 bonding electron = 7
FC = 5 - 6 = -1
(b) In Structure B
Left-hand N:
VE = 5
BE = 2 lone pairs (4)+ 2 bonding electrons = 4 + 2 = 6
FC = 5 - 6 = -1.
Central N:
VE = 5
BE = 4
FC = 5 - 4 = +1
On O:
VE = 6
BE = 2 lone pairs(4) + 2 bonding electrons = 6
FC = 6 - 6 = 0
(c) In Structure C
Left-hand N:
VE = 5
BE = 3 lone pairs (6)+ 1 bonding electrons = 6 + 1 = 7
FC = 5 - 7 = -2.
Central N:
VE = 5
BE = 4
FC = 5 - 4 = +1
On O:
VE = 6
BE = 1 lone pair(2) + 3 bonding electrons = 5
FC = 6 - 5 = +1
Keep trying you will get if you keep trying ok