Answer:
The correct movement would be -
1. Water - into solution A.
2. NaCl - into solution A.
3. glucose - into Solution B.
4. Albumin - neither.
Explanation:
All the substances are separated by the semipermeable membrane and the semipermeable membrane allows the only small molecule to pass through it. So the movement of the given substance would be -
1. Water - into solution A.
Water molecules are small and can easily pass through the semipermeable membrane as it is given that the solution b has low solute concentration and solution A has high solute concentration. It is known that the movement of the solvent always takes place from low solute concentration to high so the movement of water will be into solution A.
2. NaCl - into solution A.
The movement of small ionic molecule NaCl is always from high to low concentration as it is given that solution B has high concentration than solution A so movement will take place into solution A.
3. glucose - into Solution B.
It is also a small molecule and moves from the high glucose region to the low glucose concentration region, in solution A the concentration of glucose is high than solution B so movement would be into solution B.
4. Albumin - neither.
Albumin is a protein which is macromolecule and large in size to pass through the semipermeable membrane so, albumin move neither solution A nor solution B.
The answer is the first option. Gas burning in an engine is an example of a chemical change. Chemical change is a change where the substance changes in identity or form new substances after undergoing a process. In this case, the gas reacts with oxygen forming combustion products, commonly carbon dioxide and water.
Answer:
The answer to your question is D.
Explanation:
The latitudes near the equator receives the most direct solar energy.
Hope this helps :)
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
