<h3>Answer:<u><em>What organism is responsible for the cycling of nitrogen?</em></u></h3><h3><em><u /></em></h3><h3>Explanation:<u><em>Bacteria</em></u></h3><h3><u><em>Bacteria play a key role in the nitrogen cycle.</em></u></h3><h3><u><em>Some species of nitrogen-fixing bacteria are free-living in soil or water, while others are beneficial symbionts that live inside of plants.</em></u></h3><h3><em><u /></em></h3>
Answer:
The exact result is 82.5 min (close to c. 80 min)
Explanation:
Hello,
In the following picture you'll find the numerical procedure for this exercise.
- Take into account that the initial concentration and the velocity constant remain constant for the two mentioned scenarios.
Best regards.
Answer:
37S
Explanation:
Radioactivity is the spontaneous emission of particles and / or electromagnetic radiation by unstable atomic nuclei leading to their disintegration.
We have two main types of radioactivity: radioactive decay and artificial transmutation.
In radioactive decay ( natural radioactivity ), a naturally occurring radioactive element like Uranium-238 disintegrates or decays into more stable isotopes with the emission of particles and/or radiation.
23892U = 23490Th + 42He
Artificial transmutation is the collision of two particles where one particle captures the other used to bombard it. There is subsequent production of isotopes similar or different from the bombarded particle. Neutrons, alpha particles ( helium nucleus ), electrons, protons can be used to bombard elements.
147N + 42He = 178O + 11P
For the above question which is artificial transmutation, the reaction equation is
4018Ar + 10n = 3716S + 42He
So, the neutron capture by Argon-40 will produce a radioisotope Sulphur-37 with the emission of an alpha particle.
Answer:
Br2 (l) + 2e- ---------> 2Br- (aq) E° = 1.08 V cathode
Cu2+ (aq) + e- --------->Cu+ (aq) E° = 0.15 V anode
Explanation:
We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.
The overall reaction equation is;
2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)
E°cell= E°cathode - E°anode
E°cathode= 1.08 V
E°anode= 0.15V
E°cell = 1.08-0.15 = 0.93 V
But
∆G°= -nFE°cell
n= 2, F=96500C, E°cell= 0.93V
∆G° = -(2× 96500× 0.93)
∆G= -179490 J
But;
∆G = -RTlnK
R=8.314 JK-1
T= 25+273= 298K
Kc= the unknown
∆G° = -179490 J
Substituting values and making lnK the subject of the formula
lnK= ∆G/-RT
lnK= -( -179490/8.314 × 298)
lnK= 72.45
K= e^72.45
K= 2.91×10^31