Answer:
No.of moles of C is , n = mass/molar mass = 75.46 g / 12 (g/mol) = 6.3 moles No.of moles of H is , n' = mass/molar mass = 4.43 g / 1.0(g/mol) = 4.43 moles No.of moles of O is , n'' = mass/molar mass = 20.10 g / 16(g/mol) =1.25 moles Ratio to the no.of moles of C,H& O is 6.3 : 4.43 : 1.25 In the simple integer ratio is ( 6.3/1.25) : ( 4.43/1.25) : (1.25/1.25) 5.04 :3.5 : 1
Explanation:
Answer:
Reproductive cells(also known as sex cells) are gametes.
Explanation:
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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:
C）52g KCl in 100g water at 80°C
Explanation:
A saturated solution is one that contains as much solute as it can dissolve in the presence of excess solute at that particular temperature.
A solutibility curve is a graph that shows the variability with temperature of the solubility of a solute in a given solvent. A solutibility curve can provide information of whether a solution formed frommthe solute and solvent are saturated or not at a given temperature.
From the solubility curve in the attachment below:
A) A saturated solution of NH₄Cl will contain about 52 g solute per 100 g sat 50 °C. Thus, a solution of 40 g NH₄Cl in 100 g water at 50 °C is an unsaturated solution.
B) A saturated solution of SO₂ at 10°C will contain about 70 g of solute in 100 g of water. Thus a solution of 2g SO₂ in 100g water at 10°C is an unsaturated solution.
C) A saturated solution of KCl at 80 °C will contain about 52 g of solute in 100 g of water. Thus, a solution of 52g KCl in 100g water at 80°C is a saturated solution.
D) A saturated solution of Kl at 20 °C will contain about 145 g of solute in 100 g of water. Thus, a solution of 120g KI in 100g water at 20°C is an unsaturated solution.
Answer:
Therefore 500 ml of solution have 0.25 mol of NaCl .. = 14.61 g (ans.)
Explanation:
