Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.
The single most important chemical weathering agent is Carbon dioxide.
Weathering refers to the process that change the physical and chemical character of rock at or near the surface. Weathering has a dramatic impact on the composition of Earth's atmosphere. Chemical weathering removes carbon dioxide from the atmosphere, allowing it to be transformed into limestone and stored in the crust. Without chemical weathering, the elevated levels of carbon dioxide in the atmosphere would have long made Earth too hot to sustain life.
Answer:
Pu-239
Explanation:
Beta decay moves the element which undergoes the decay one place to the right in the periodic table since to conserve charge and being beta radiations an electron we convert a neutron into a proton and an electron. In neutron capture we increase the atomic mas by one unit. We that in mind, lets solve the question:
U-238 + ₁⁰ n ⇒ U-239 ⇒ Np -239 + ₋₁⁰β ⇒ Pu-239 + ₋₁⁰β
Answer:
M Na2SO4 sln = 0.8448 M
Explanation:
∴ mass Na2SO4 = 3.00 g
∴ volume soln = 25 mL = 0.025 L
∴ molar mass Na2SO4 = 142.04 g/mol
⇒ mol Na2SO4 = (3.00 g)*(mol/142.04 g) = 0.02112 mol
⇒ M Na2SO4 sln = (0.02112 mol/0.025 L ) = 0.8448 M
If 30 grams of KCl is dissolved at 10°C, 14 g of KCl should be added to make a saturated solution at 60 °C.
<h3>What is a saturated solution?</h3>
A saturated solution is a solution in which there is so much solute that if there was any more, it would not dissolve. Its concentration is the same as the solubility at that temperature.
- Step 1. Calculate the mass of water.
At 10 °C, the solubility is 31.2 g KCl/100 g H₂O.
30 g KCl × 100 g H₂O/31.2 g KCl = 96 g H₂O
- Step 2. Calculate the mass of KCl required to prepare a saturated solution at 60 °C.
At 60 °C, the solubility is 45.8 g KCl/100 g H₂O.
96 g H₂O × 45.8 g KCl/100 g H₂O = 44 g KCl
- Step 3. Calculate the mass of KCl that must be added.
44 g - 30 g = 14 g
If 30 grams of KCl is dissolved at 10°C, 14 g of KCl should be added to make a saturated solution at 60 °C.
Learn more about saturated solutions here: brainly.com/question/24564260
