Explanation:
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Answer:
The correct answer is AMP+H2O→ Adenosine + pi
Explanation:
The above reaction is least energetic because there is no phosphoanhydride bond present with adenosine mono phosphate.Phospho anhydride bond is an energy rich bond.
As a result hydrolysis of AMP generates very little amount of energy in comparison to the hydrolysis of ATP and ADP.
Answer:
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Explanation: genetically modified organisms (GMOs) are being made by inserting a gene from an external source such as viruses, bacteria, animals or plants (etc) into foods. Genetically modified foods seem to be not as healthy as unmodified foods, according to research. also many people are against it due to the fact that some people go overboard and add too many genetically modified organisms, which causes health risks and unfortunately harms people. sometimes genetically modified food contains toxins and antibiotics. srry i couldnt find much on the process but this is based off my background knowledge so yeah
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
<span>1. Translate, predict the products, and balance the equation above.
Li + Cu(NO3)2 = Li(NO3)2 + Cu
2. How many particles of lithium are needed to produce 125 g of copper?
125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles
3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?
</span>4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = 1043.77 grams Li(NO3)2
