Answer:
Explanation:
1) Vanadium (IV) → V⁺⁴
Carbonate → CO₃⁻²
So , Vanadium (IV) Carbonate = V₂(CO₃)₄ or V(CO₃)₂
2) Tin (II) = Sn⁺²
Nitrite = NO₂⁻
So, Tin (II) Nitrate = Sn(NO₂)₂
3) Cobalt (III) = Co⁺³
Oxide = O⁻²
So , Cobalt (III) Oxide = Co₂O₃
4) Titanium (II) = Tn⁺²
Acetate = CH₃COO⁻
So , Titanium (II) Acetate = Tn(CH₃COO)₂ or Tn(C₂H₃O₂)₂
5) Vanadium (V) = V⁺⁵
Sulfide = S⁻²
So , Vanadium (V) Sulfide = V₂S₅
6) Chromium (III) = Cr⁺³
Hydroxide = OH⁻
So , Chromium (III) Hydroxide = Cr(OH)₃
7) Lithium = Li⁺
Iodide = I⁻
So , Lithium Iodide = LiI
8) Lead (II) = Pb⁺²
Nitride = N⁻³
So , Lead (II) Nitride = Pb₃N₂
9) Silver = Ag⁺
Bromide = Br⁻
So , Silver Bromide = AgBr
When an electron quickly occupies an strength state increased than its ground state, it is in an excited state. An electron can end up excited if it is given greater energy, such as if it absorbs a photon, or packet of light, or collides with a close by atom or particle.
Answer:
Rate of the reaction is 0.2593 M/s
-0.5186 M/s is the rate of the loss of ozone.
Explanation:
The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.
Rate of formation of oxygen : 
Rate of the reaction(R) =![\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B2%7D%5Cfrac%7Bd%5BO_3%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![R=\frac{1}{3}\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
Rate of formation of oxygen=3 × (R)
Rate of the reaction(R): 
Rate of the reaction is 0.2593 M/s
Rate of disappearance of the ozone:
![R=-\frac{1}{2}\frac{d[O_3]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BO_3%5D%7D%7Bdt%7D)
![\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_3%5D%7D%7Bdt%7D%3D-2%5Ctimes%20R%3D-2%5Ctimes%200.2593%5Ctimes%20M%2Fs%3D-0.5186M%2Fs)
-0.5186 M/s is the rate of the loss of ozone.
Answer:
Option 2= Glucose
Explanation:
Cell membrane is made up of two phospholipid layers and each contain phosphate head and fatty acid or lipid tails. the head is present between the outer and inner boundaries and tail is present in between. The small non- polar molecules can pass the membrane through simple diffusion. This lipid tail restrict the passage of polar molecules including water soluble substances like glucose. However, transmembranes are present that allow the molecules to inter that are blocked by the tails.
Facilitated diffusion:
it is a type of diffusion in which caries protein without using the cellular energy shuttle the molecules to the cell membrane. Glucose is bind on the carrier protein ,change the shape and transport it from one to another side of membrane. In order to absorb the glucose red blood cells use this kind of diffusion.
Primary active transport:
The cells that are present along small intestine use this type of transport to pump the glucose inside the cell. The primary active transport require energy to transport the glucose inside.
Secondary active transport:
It is another method of transport of glucose into the cell. This method can not use ATP but it is based on concentration gradient of the sodium that provide electro chemical energy for the glucose transport.
Move the decimal place to the left 3 digits.
0.125
