Answer:
In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure 2.2.2 ). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
Explanation:
A planetary surface is where the solid (or liquid) material of the outer crust on certain types of astronomical objects contacts the atmosphere or outer space. Planetary surfaces are found on solid objects of planetary mass, including terrestrial planets (including Earth), dwarf planets, natural satellites, planetesimals and many other small Solar System bodies (SSSBs).[1][2][3] The study of planetary surfaces is a field of planetary geology known as surface geology, but also a focus of a number of fields including planetary cartography, topography, geomorphology, atmospheric sciences, and astronomy. Land (or ground) is the term given to non-liquid planetary surfaces. The term landing is used to describe the collision of an object with a planetary surface and is usually at a velocity in which the object can remain intact and remain attached.
In differentiated bodies, the surface is where the crust meets the planetary boundary layer. Anything below this is regarded as being sub-surface or sub-marine. Most bodies more massive than super-Earths, including stars and gas giants, as well as smaller gas dwarfs, transition contiguously between phases, including gas, liquid, and solid. As such, they are generally regarded as lacking surfaces.
Planetary surfaces and surface life are of particular interest to humans as it is the primary habitat of the species, which has evolved to move over land and breathe air. Human space exploration and space colonization therefore focuses heavily on them. Humans have only directly explored the surface of Earth and the Moon. The vast distances and complexities of space makes direct exploration of even near-Earth objects dangerous and expensive. As such, all other exploration has been indirect via space probes.
Indirect observations by flyby or orbit currently provide insufficient information to confirm the composition and properties of planetary surfaces. Much of what is known is from the use of techniques such as astronomical spectroscopy and sample return. Lander spacecraft have explored the surfaces of planets Mars and Venus. Mars is the only other planet to have had its surface explored by a mobile surface probe (rover). Titan is the only non-planetary object of planetary mass to have been explored by lander. Landers have explored several smaller bodies including 433 Eros (2001), 25143 Itokawa (2005), Tempel 1 (2005), 67P/Churyumov–Gerasimenko (2014), 162173 Ryugu (2018) and 101955 Bennu (2020). Surface samples have been collected from the Moon (returned 1969), 25143 Itokawa (returned 2010), 162173 Ryugu and 101955 Bennu.
<h2>Answer: C) 1s²2s²2p⁶</h2>
<h3>Explanation:</h3>
A noble gas has 8 electrons between the p and s orbitals of the outer shell. Helium is the exception because it only has two electrons.
<h3> ∴ 1s²2s²2p⁶ is the noble gas (neon)</h3>
Volume of Hydrogen V1 = 351mL
Temperature T1 = 20 = 20 + 273 = 293 K
Temperature T2 = 38 = 38 + 273 = 311 K
We have V1 x T2 = V2 x T1
So V2 = (V1 x T2) / T1 = (351 x 311) / 293 = 372.56
Volume at 38 C = 373 ml
Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
=14-56.17
=8.823
The volume of the 
= 40.00 ml
convert into the liter= 0.040L
The value of the concentrated =0.10 M
The volume of the 
= 50.00 ml
convert into the liter= 0.050L
The value of concentrated = 0.10 M
The volume of the 
= 30 ml
convert into the liter= 0.030L
The value of concentrated 
=0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of 
= 
calculating the new concentrated value of 
= 
calculating the new concentrated value of 
when 1 mol produced 2 mols 
so, 0.0125 in produced:
create the ICE table:
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83:
![P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769](https://tex.z-dn.net/?f=P%5E%7BH%7D%3D%20p%5E%7Bkb%7D%7C%2B%20%5Clog%5Cfrac%7B%5BNH_4%5E%7B%2B%7D%5D%7D%7B%5BNH_3%5D%7D%5C%5C%5C%5C8.83%3Dp%5E%7Bkb%7D%2B%5Clog%5Cfrac%7B0.0667%7D%7B8.3%20%5Ctimes%2010%5E%7B-3%7D%7D%5C%5C%5C%5Cp%5E%7Bkb%7D%3D8.83-0.9069%5C%5C%5C%5C%20%5C%20%5C%20%5C%20%3D7.7231%20%5C%5C%5C%5C%5C%20The%20P%5E%7Bkb%7D%20%5C%20for%20%5C%20NH_3%20%5C%20is%20%3D7.7231%5C%5C%5C%5C%5C%20The%20P%5E%7Bkb%7D%20%5C%20for%20N%5E%7B%2B%7DH_4%3D14-7.7231%5C%5C%5C%5C%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%3D6.2769)
