Answer:
It can be concluded that the third step of the reaction is very fast, in this way, it does not contribute to the rate law
Explanation:
Please, observe the solution in the attached Word document.
I’m writing this equation by memory, so I hope I’m correct. It’s been about four months since we used in in my chem class:
(P-(n^2•a)/V^2)(V-nb)=nRT
Plugging in values given:
(P-(1•1.35)/(1.42^2))(1.42-(1•0.0322))=(1)(0.0821)(300)
(P-(1.35/2.016))(1.42-0.0322)=24.63
(P-(1.35/2.016))=17.75
P=18.42 atm
The pressure exerted by the Argon would be 18.42 atmospheres.
<span>The
energy that was produced is called, heat combustion. It is an energy released
in the form of heat when chemicals are mixed. An explosion of light and sound
are the common characteristics of heat combustion. The chemical reaction takes
place because of the presence of oxygen and hydrocarbon or organic molecule
substances that when mixed form carbon dioxide and water which then releases
heat and explosive characteristics.</span>
Answer:
The correct answer is - option D. (check image)
Explanation:
Alkynes and alkenes both decolorized bromine in carbon tetrachloride. The absorption of the IR at about 3300 cm-1 for the X here that are found in the terminal alkynes absorption range only. In presence of excess hydrogen and a nickel catalyst, x gives the 2-methyl pentane.
The most likely structure for X is: CH3-CH3-ch-CH2-C≡CH
Answer:
2.1x10⁹ years
Explanation:
U-238 is a radioactive substance, which decays in radioactive particles. It means that this substance will lose mass, and will form another compound, the Pb-206.
The time need for a compound loses half of its mass is called half-life, and knowing the initial mass (mi) and the final mass (m) the number of half-lives passed (n) can be found by:
m = mi/2ⁿ
The mass of Pb-206 will be the mass that was lost by U-238, so it will be mi - m. Thus, the mass ration can be expressed as:
(mi-m)/m = 0.337/1
mi - m = 0.337m
mi = 1.337m
Substituing mi in the expression of half-life:
m = 1.337m/2ⁿ
2ⁿ = 1.337m/m
2ⁿ = 1.337
ln(2ⁿ) = ln(1.337)
n*ln(2) = ln(1.337)
n = ln(1.337)/ln2
n = 0.4190
The time passed (t), or the age of the sample, is the half-life time multiplied by n:
t = 4.5x10⁹ * 0.4190
t = 1.88x10⁹ ≅ 2.1x10⁹ years
