Chloroethane is the answer
Explanation:
<h3>PLA is a polyester produced by fermentation under controlled conditions of a carbohydrate source like corn starch or sugarcane. ... The starch is then mixed with acid or enzymes and heated. This process “breaks” starch into dextrose (D-glucose), or corn sugar.</h3>
<h3>PLA is a polymer made from high levels of polylactic acid molecules. </h3><h3>For PLA to biodegrade, you must break up the polymer by adding </h3><h3>water to it (a process known as hydrolyzing). Heat and moisture are required for hydrolyzing to occur.</h3>
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Answer:
1.0 ° C
Explanation:
The molar mass for Sodium Nitrate NaNO₃ = (23+14+(16×3)) = 85
Number of moles of NaNO₃ = mass of NaNO₃ /molar mass of NaNO₃
⇒ 17/85 = 1.38 moles
Since 1 mole of NaNO₃ dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed.
when 1.38 mole of NaNO₃ dissolved in 1 cubic decimeter of water, x kJ of heat energy is absorbed..
Then; x kJ of 1.38 mole of NaNo₃ = 1.38 × 40 kJ =55.2 kJ of heat absorbed.
Using the relation : Q = mcΔT to determine the temperature drop ; we get:
55.2 = 17 × 4 (ΔT)
55.2 = 68 ΔT
ΔT= 0.8 ° C
ΔT ≅ 1.0 ° C
Therefore, the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water is 1.0 ° C
Answer: Yes, 
is a strong acid.
acid = , conjugate base = 
, base = 
, conjugate acid = 
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
Yes is a strong acid as it completely dissociates in water to give 
ions.
For the given chemical equation:
Here, is loosing a proton, thus it is considered as an acid and after losing a proton, it forms which is a conjugate base.
And, is gaining a proton, thus it is considered as a base and after gaining a proton, it forms which is a conjugate acid.
Thus acid =
conjugate base =
base =
conjugate acid = .
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.
