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3 years ago

Write the products of photosystem i and photosystem ii

Chemistry

2 answers:

lisabon 2012 [21]3 years ago

8 0

Answer:

Explanation:

Photosystems are generally involved in photosynthesis, they assist in carrying out basic photochemistry (absorption of light and transfer of energy) during photosynthesis.

Photosystem I (or PSI) produces high energy compounds; ATP and NADPH from light energy absorbed during the light reactions of photosynthesis in algae, plants and some bacteria.

Photosystem II is the first membrane protein complex in organisms that produce oxygen. This oxygen oxidizes two molecules of water (in the presence of light) to produce molecular oxygen.

Margaret [11]3 years ago

3 0

ATP,O2 and NADPH are the products. H2O,NADP,ADP and Pi are the reactants. acts as an electron carrier between the cytochrome b6f and photosystem 1 (PS1) complexes in the photosynthetic electron-transfer chain.

Photosystem II (or water-plastoquinone oxidoreductase) is the first protein complex in the light-dependent reactions of oxygenic photosynthesis. It is located in the thylakoid membrane of plants, algae, and cyanobacteria.

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Which forces involve nonpolar molecules?

Elanso [62]

The answer is London dispersin forces and dipole-induced dipole forces.

The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction. This force is found in any compound and is the weakest atraction force between atoms or molecules.

Those temporay dipoles are not like the dipoles that form the polar molecules, because the polar molecules are the result of permanent dipoles.

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4 years ago

Read 2 more answers

What is the specific heat capacity of propane in J/gC?

Anna007 [38]

Answer:

At ambient pressure and temperature the isobaric specific heat, CP, of gaseous propane is 1.68 kJ/kg K or 0.402 Btu/lb °F = cal/g K, while the isochoric specific heat, CV, is 1.48 kJ/kg K or 0.353 Btu/lb °F = cal/g K.

Explanation:

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3 years ago

Researchers can use solvents to extract, or break apart, components of a cell or tissue to separate and/or isolate components. I

Tems11 [23]

The cell membrane which is more fat permeable will be present at the end of the process.

Solvent extraction is a very important method if separating mixtures. The principle of solvent extraction is based on the idea of like dissolves like. A substance dissolves in the component of the system in which it is most soluble.

If a cell extract is dissolved in a nonpolar solvent, the cell membrane which is mostly permeable to nonpolar molecules will dissolve most in the solvent.

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2 years ago

At the end of the isomerization reaction, what chemical is used to quench the residual bromine?At the end of the isomerization r

valina [46]

Answer: Cyclohexene

Explanation:

Cyclohexane belongs to the Alkenes family. Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom get attached to each carbon. The bromine loses its original red-brown color to give a colorless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed. When bromine is added to cyclohexane in the dark room, there won't be any reaction. If the mixture is exposed to light however, free bromine radicals are generated. In this condition, polybrominated products can be produced as well.

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4 years ago

A solution of fructose, C6H12O6, a sugar found in many fruits, is made by dissolving 34.0 g of fructose in 1.00 kg of water. Wha

Mashcka [7]

Answer:

Molality → 0.188 m

Mole fraction of fructose → 0.00337

Mass percent of fructose in solution → 3.29 %

Molarity → 0.183 M

Explanation:

Solute → 34 g of fructose

Solvent → 1000 g of water

Solution → 1000 g of water + 34 g of fructose = 1034 g of solution.

We take account density to calculate, the solution's density

1.0078 g/mL = 1034 g / mL

1034 g / 1.0078 g/mL = 1026 mL

Molal concentration → moles of solute in 1kg of solvent

Moles of fructose = mass of fructose / molar mass

34 g/ 180g/mol = 0.188 mol

0.188 mol/1kg = 0.188 m

Mole fraction of fructose = Moles of fructose / Total moles

We determine the moles of water

Moles of water = 1000 g / 18 g = 55.5 mol

Total moles = moles of fructose + moles of water

0.188 mol + 55.5 mol = 55.743 mol

0.188 mol / 55.743 mol = 0.00337

Mass percent = mass of fructose in 100 g of solution

(Mass of fructose / Total mass ) . 100 = (34 g /1034 g) . 100 = 3.29 %

Molarity = Moles of solute in 1L of solution

We can also say mmol of solute in 1 mL of solution

0.188 mol of fructose = 188 mmol of fructose

Molarity = 188 mmol / 1026 mL of solution = 0.183 M

8 0

3 years ago