Answer:
The final temperature of sulfur dioxide gas is 215.43 C
Explanation:
Gay Lussac's Law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that if the temperature increases the pressure increases, while if the temperature decreases the pressure decreases. In other words, the pressure and temperature are directly proportional quantities.
Mathematically, the Gay-Lussac law states that, when a gas undergoes a transformation at constant volume, the quotient of the pressure exerted by the temperature of the gas remains constant:
Assuming you have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2, and it will be true:
The reference temperature is the absolute temperature (in degrees Kelvin)
In this case:
- P1= 0.450 atm
- T1= 20 C= 293.15 K (being 0 C= 273.15 K)
- P2=0.750 atm
- T2= ?
Replacing:
Solving:
T2=488.58 K
Being 273.15 K= 0 C, then 488.58 K= 215.43 C
<u><em>The final temperature of sulfur dioxide gas is 215.43 C</em></u>
Answer: (Structure attached).
Explanation:
This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.
The aromatic ring in p-fluoroanisole has two sustituents, an <u>halogen</u> and a <u>methoxy group</u>, which are <em>ortho-para</em> directing substituents.
Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using <u>sulfur trioxide as an electrophile</u> (very reactive).
The reaction occurs in three steps:
- The attack on the electrophile forms the sigma complex.
- The loss of a proton regenerates an aromatic ring.
- The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).
Normally, a mixture of <em>ortho-para</em> substituted products would be obtained. However, since both <em>para</em> positions are occupied, only the <em>ortho </em>substituted product is obtained here.
Explanation:
from the graph study about oxygen content of Earth's atmosphere, we can understand that
1)
4 billions year ago = None, 3 billions year ago = Cyanobacteria and Archaea , 2 and 1 billions year ago = Bacteria and Green algae , 500 Ma = invertebrate fossils started to existence. Early land plants came in to existence around 398 MA that is Devonian. Dinosaurs are came in to existence during Triassic and Jurassic that is around 251 Ma. Man and animals are recent organism came under Holocene that is 11000 years ago.
2)
The first cells on the earth are anaerobic microorganisms, as the CO2 level is too high they survive by using CO2.
3)
Starting around 2.7 billion years ago, photosynthesis by Cyanobacteria and later plants , pumped “OXYGEN” in to the atmosphere. This caused the decline of anaerobic bacteria and allows the diversification of animals as seen in “CAMBRIAN” around 500 millions year ago.
Early vascular plants “CAPTURED” CO2 starting before the Carboniferous period that began around 350 millions year.Leading to lower temperatures and allowing and allowing the seed plants to outcompetes seedless plants.
Modern human activities has raised both “CO2 and METHANE” level in the atmosphere to over leading to higher temperature and extinction of other species.
Answer:
Um I can help but not on zo.om. I can help u on here just post ur question and I will answer it
The rate of a reaction would be one-fourth.
<h3>Further explanation</h3>
Given
Rate law-r₁ = k [NO]²[H2]
Required
The rate of a reaction
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
Can be formulated:
Reaction: aA ---> bB
or
The concentration of NO were halved, so the rate :
![\tt r_2=k[\dfrac{1}{2}No]^2[H_2]\\\\r_2=\dfrac{1}{4}k.[No]^2[H_2]\\\\r_2=\dfrac{1}{4}r_1](https://tex.z-dn.net/?f=%5Ctt%20r_2%3Dk%5B%5Cdfrac%7B1%7D%7B2%7DNo%5D%5E2%5BH_2%5D%5C%5C%5C%5Cr_2%3D%5Cdfrac%7B1%7D%7B4%7Dk.%5BNo%5D%5E2%5BH_2%5D%5C%5C%5C%5Cr_2%3D%5Cdfrac%7B1%7D%7B4%7Dr_1)
