None of the questions asked can be answered completely from the graph provided (GHG emissions: Direct, indirect and total Vs Year)
Reason:
1) Question A:<span>What caused a drop in GHG emissions around 2009?. This questions in pointing towards reason for drop of GHG emission around 2009. From the graph, it can be seen that there is a drop in GHG emission around 2009. However, information for reason for this drop is not available in graph.
2) Question B: </span>Did GHG emissions cause the melting of Arctic glaciers?. As mentioned earlier, the graph plotted provides information of GHG emissions: Vs Year. Information related to impact of GHG on environment is not available in graph.
3) Question C: <span>How much methane was emitted by homes between 1990 and 2000?. Graph provides information of direct and indirect emission for GHG. However, it lacks information about emission from residential or industrial sources.
4) </span>Question D: <span>Does industrial equipment release gases other than greenhouse gases?: Present study doesnot cover type of gases emitted from industrial equipment.
5) </span>Question E: <span>Which types of industries were included in the study?: Present graph has not specific information related to industries. </span>
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after 
is 0.00345 mol/L.
Explanation:
Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t =
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after is 0.00345 mol/L.
Answer:
Molarity =5.32 M
Explanation:
Given data:
Mass of glucose = 239 g
Volume = 250 mL (250 /1000 = 0.25 L)
Molarity = ?
Solution;
Formula:
Molarity = number of moles / volume in litter
Number of moles:
Number of moles = mass/ molar mass
Number of moles = 239 g / 180.2 g/mol
Number of moles = 1.33 mol
Molarity:
Molarity = number of moles / volume in litter
Molarity = 1.33 mol / 0.25 L
Molarity =5.32 M
Divide both sides to get 104 and that’s your answer
The answer is “Heat Conduction”
