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Katena32 [7]
2 years ago
14

Which piece of the planetary object data could be used to decide if there actually is a solid surface to land on?

Chemistry
1 answer:
zloy xaker [14]2 years ago
3 0
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.
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Ilia_Sergeevich [38]
I think the answer is D

8 0
2 years ago
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Which statments regarding the henderson-hasselbalch equation are true?
ziro4ka [17]

Complete question is;

Which statements regarding the Henderson-Hasselbalch equation are true?

1. If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined.

2. At pH = pKa for an acid, [conjugate base] = [acid] in solution.

3. At pH > pKa for an acid, the acid will be mostly ionized.

4. At pH < pKa for an acid, the acid will be mostly ionized.

A. All of the listed statements are true. B. 1, 2, and 3 are true.

C. 2, 3, and 4 are true.

D. 1, 2, and 4 are true.

Answer:

B. 1, 2, and 3 are true.

Explanation:

The formula for the Henderson-Hasselbalch equation is:

pH = pka + log₁₀([A^(-)]/[HA])

Where;

PH is acidity of solution

ka is acid dissociation constant

A^(-) is concentration of conjugate base

HA is concentration of Acid

- For statement 1; If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined;

pH = pka + log₁₀([A^(-)]/[HA])

pH - pka = log₁₀([A^(-)]/[HA])

10^(pH - pka) = ([A^(-)]/[HA])

Since we can find the ratio as seen, then the statement is true

- For statement 2: At pH = pKa for an acid, [conjugate base] = [acid] in solution;

We will substitute pH for pKa;

pH = pH + log₁₀([A^(-)]/[HA])

This give;

0 = log₁₀([A^(-)]/[HA])

10^(0) = [A^(-)]/[HA]

1 = [A^(-)]/[HA]

Thus; [A^(-)] = [HA]

Thus, the statement is true

- For statement 3: At pH > pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is greater than 0 and thus;

10^(pH - pKa) is greater than 1.

Thus;

[A^(-)]/[HA] > 1

[A^(-)] > [HA]

So more acid is ionized than base.

So the statement is true.

- For statement 4: At pH < pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is less than 0 and thus;

10^(pH - pKa) is less than 1.

Thus;

[A^(-)]/[HA] < 1

[A^(-)] < [HA]

So we have more base ionized than acid. So statement is false

7 0
3 years ago
Assume that you have an unknown consisting of an aqueous solution of a salt that contains one of the ions listed above. Which io
bixtya [17]

Answer:

Check the answers in the explanation

Explanation:

The following are the answers

(1) if the solution is colourless :  The answer is   B Cr207

(2) If the solution gives no apparent reaction with dilute  hydrochloric acid, : The answer is  A CO3

(3) if  No odour can be detected when a sample of the solution is added drop by drop to a warm solution of sodium hydroxide  : The answer is  C NH4

(4) If No precipitate is formed when a dilute solution of H2SO4 is added to a sample of the solution : The answer is  D ba

4 0
3 years ago
According to the following reaction, how many grams of hydrogen peroxide (
Yuri [45]

Answer:

12.4 g

Explanation:

Let's consider the following balanced equation.

H₂O₂(aq) → H₂O(l) + 0.5 O₂(g)

The molar ratio of H₂O₂ to O₂ is 1:0.5. The moles of H₂O₂ required to form 0.182 moles of O₂ are:

0.182 mol O₂ × (1 mol H₂O₂/ 0.5 mol O₂) = 0.364 mol H₂O₂

The molar mass of H₂O₂ is 34.01 g/mol. The mass of H₂O₂ corresponding to 0.364 moles is:

0.364 mol × 34.01 g/mol = 12.4 g

5 0
3 years ago
How does solidification of alloys differ from solidification of pure metals?.
jarptica [38.1K]

Answer:

Solidifications of alloys occurs over a range of temperatures whereas solidification of a pure metal will take place at a constant temperature

6 0
2 years ago
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