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aniked [119]
3 years ago
9

How are the excited electrons from stage 1 used in stage 2 of photosynthesis? A. Splitting of water b. Carbon fixation c. Format

ion of ADP d. Formation of NADPH
Chemistry
1 answer:
Gnom [1K]3 years ago
6 0

Answer;

D. Formation of NADPH

Explanation;

During the process of photosynthesis light is converted to chemical energy.

During stage 1, Excited electrons that leave the chlorophyll pigments in stage 1 are used. Excited electrons are passed through proteins in the thylakoid membrane like a ball being passed from person to person. H2o molecules split and turn into H+ molecules and O2 gas, then pigments take the electrons from the split water molecules and release O2 into the atmosphere.

During stage 2; electrons from 1st cluster pump H+ ions into the thylakoid membrane through the hydrogen ion pump (protein membrane. Higher concentration of H+ ions inside the thylakoid membrane compared to outside. H+ ions diffuse out of the thylakoid throught the ATP synthase (enzyme in membrane). ATP synthase catalyzes the reaction ADP + P ---->ATP.

Stage 2; Excited electrons + H+ ions + NADP+= NADPH

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A solution contains some or all of the ions Cu2+,Al3+, K+,Ca2+, Ba2+,Pb2+, and NH4+. The following tests were performed, in orde
xeze [42]

Answer:

See below explanation

Explanation:

When having a mixture of metals in solution, you may perform an analytical study (using selective chemical conditions), that may help you to determine whether a metal (cation) is present or not

Using selective analytes (or conditions), leads to consecutive precipitations, until most of the cations are separated in precipitates

With this technique, you may identify metals in different groups, each group will have its analyte (or condition), which will help to have a different precipitate:

- Group I: Ag⁺, Pb⁺², Hg⁺²;  Analyte: HCL ; Precipitate: AgCl (white) , PbCl₂, HgCl₂

- Group II: As⁺³ , Bi⁺³, Cd⁺², Cu⁺² , Sb⁺³, Sn⁺² ; Analyte: H₂S (g) with HCL ; Precipitate: As₂S₃ , Bi₂S₃ , CdS (yellow) , CuS (black), Sb₂S₃, SnS

- Group III: Co⁺², Fe⁺², Fe⁺³, Mn⁺², Ni⁺², Zn⁺², Al⁺³, Cr⁺³; Analyte: NaOH or NH₃ with (NH₄)₂S (ac) ; Precipitate: CoS (black) , FeS, MnS , NiS (black), ZnS (white) , Al(OH)₃ (white), Cr(OH)₃  

- Group IV: Mg⁺², Ca⁺², Sr⁺², Ba⁺²; Analyte: Na₂CO₃ (ac) or (NH₄)₂HPO₄ (ac); Precipitate: respective carbonate or phosphate MgCO₃/MgHPO₄, CaCO₃/CaHPO₄ , SrCO₃/SrHPO₄, BaCO₃/BaHPO₄

- Group V: Li⁺, K⁺, Na⁺, Rb⁺, Cs⁺, NH₄⁺ ; will remain all in final solution

According to the original statement:

A solution contains one or more of the following: Cu⁺², Al⁺³, K⁺, Ca⁺², Ba⁺², Pb⁺², NH₄⁺

1) Addition on HCl 6M produces no change: we can say the sample does not contain Pb⁺² (group I)

2) Addition of H₂S with 0.2 M HCL produced a black solid: we could say sample contains Cu⁺²(group II)

3) Addition of (NH₄)₂HPO₄ in NH₃ produces no reaction: we could say we don´t have Ca⁺² and /or Ba⁺²  (group IV)

4) The final supernatant, when heated produced a purple flame: in the final solution, we have K⁺ (group V), which produces a purple flame (based on its characteristic emission spectrum when subjected to flame)

This analysis will be inconclusive for NH₄⁺ (according to above describe technique)

6 0
3 years ago
Molecular iodine, I2(g), dissociates into iodine atoms at 625 K with a first-order rate constant of 0.271 s−1 .
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4 0
2 years ago
Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T
Ostrovityanka [42]

Explanation:

The given data is as follows.

          \Delta H = 286 kJ = 286 kJ \times \frac{1000 J}{1 kJ}

                            = 286000 J

 S_{H_{2}O} = 70 J/^{o}K,      S_{H_{2}} = 131 J/^{o}K

 S_{O_{2}} = 205 J/^{o}K

Hence, formula to calculate entropy change of the reaction is as follows.

          \Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)

                     = [(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]

                    = [(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]

                    = 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

             \Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}

                            = 286000 J - (163.5 J/K \times 298 K)

                            = 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

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Does litmus paper indicate if a liquid is a base?
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