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.
Answer:
350.64g
Explanation:
So first you must know that M is mol/L
Next solve the problem using dimensional analysis
2L NaCl (3 mol/L) = 6 mol NaCl
After you got the number of moles you should look at your periodic table to find the molar mass
I see that it's 58.44g/mol
Use dimensional analysis again!
6 mol (58.44g/mol) = 350.64g
Don't forget to make me brainliest!
The molar mass (atomic weight ) of sodium is 23.0 grams/mole and the molar mass of sodium azide, NaN3 , is the mass of sodium, 23.0 gram/mole added to the molar mass of three atoms of nitrogen (14.0 x 3 = 42 gram/mole) which equals 65.0 grams/mole. The percentage of sodium is 23.0 /65.0 x 100 % = 35 %
Answer:
I think its D
Explanation:
It cant be B or C bc the solute particles arent changing. And its not A because the concentration isnt constant
Answer:
See explanation below
Explanation:
To do this, we will use the following expression to calculate the [H⁺]:
[H⁺] = Kw / [OH⁻]
[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.
a) [H⁺] = 1x10⁻¹⁴ / 6x10⁻¹² = 1.67x10⁻³ M. Acidic.
b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.
c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.
d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.
e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.
f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.
g) [H⁺] = 1x10⁻¹⁴ / 5x10⁻⁵ = 2x10⁻¹⁰ M. Basic.
h) [H⁺] = 1x10⁻¹⁴ / 1x10⁻⁷ = 1x10⁻⁷ M. Neutral.
Part B.
In this part, we'll use the following expression and replace the given values:
[OH⁻] = Kw / [H⁺]
Replacing the values:
[OH⁻] = 1x10⁻¹⁴ / 5.2x10⁻⁵
[OH⁻] = 1.92x10⁻¹⁰ M
PArt C:
In this case, we will use expression of part A, and replace the given values:
[H⁺] = 1x10⁻¹⁴ / 2.7x10⁻²
[H⁺] = 3.7x10⁻¹³ M
