Answer:
C
Explanation:
it wouldn't be A because that makes no sense
theirs no pill or any type of drug that can slow it down but their is a treatment plan for it
(It could be c or d)
Answer:
C2H2O4
Explanation:
To get the molecular formula, we first get the empirical formula. This can be done by dividing the percentage compositions by the atomic masses. The percentage compositions are shown as follows :
C = 26.86%
H = 2.239%
O = 100 - ( 26.86 + 2.239) = 70.901%
We then proceed to divide by their atomic masses. Atomic mass of carbon is 12 a.m.u , H = 1 a.m.u , O = 16 a.m.u
The division is as follows:
C = 26.86/12 = 2.2383
H = 2.239/1 = 2.239
O = 70.901/16 = 4.4313
We now divide each by the smallest number I.e 2.2383
C = 2.2383/2.2383 = 1
H = 2.239/2.2383 = 1
O = 4.4313/2.2383 = 1.98 = 2
Thus, the empirical formula is CHO2.
To get the molecular formula, we use the molar mass .
(CHO2)n = 90
We add the atomic masses multiplied by n.
(12 + 1 + 2(16))n = 90
45n = 90
n = 90/45 = 2.
Thus , the molecular formula is C2H2O4
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.
<span>The purpose of washing the product with NaOH is simply to
neutralize any acid which remained or leaked after the 1st initial
separation. The NaOH base reacts with the acid to form neutralization reaction products
which are soluble in water.</span>