Explanation:
Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.
HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−
Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M
Writing the information from the ICE Table in Equation form yields
6.6×10−4=x20.3−x6.6×10−4=x20.3−x
Manipulating the equation to get everything on one side yields
0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4
Now this information is plugged into the quadratic formula to give
x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2
The quadratic formula yields that x=0.013745 and x=-0.014405
However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86
Answer:
The reactants would appear at a higher energy state than the products.
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Answer:
The reagents are 
.
Explanation:
1-Methylenecyclopentene is treated with HBr form 1-bromo-1-methylcyclopentane, which is treated with strong base ethoxide ion and forms 1-methylcyclopent-1-ene.
This alkene is treated with osmium tetraoxide in the presence of sodium bisulfite to form target product.
The chemical reaction is as follows.
