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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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Cr2o2−7(aq)+i−(aq)→cr3+(aq)+io−3(aq) (acidic solution) express your answer as a chemical equation. identify all of the phases in
Rashid [163]
<span>Answer: Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either. Start with your 2 half reactions: I- --> IO3- Cr2O72- --> 2 Cr3+ Balance O by adding H2O: I- + 3 H2O --> IO3- Cr2O72- --> 2 Cr3+ + 7H2O Balance H by adding H+: I- + 3 H2O --> IO3- + 6 H+ Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O Balance charge by adding e-: I- + 3 H2O --> IO3- + 6 H+ + 6 e- Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give: Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
4 0
3 years ago
consider the titration of hclo4 with koh. what is the ph after 17.0 ml of 0.15 m koh has been added to 15 ml of 0.20 m hclo4?
bezimeni [28]

The ph after 17.0 ml of 0.15 m Koh has been added to 15 ml of 0.20 m hclo4 is  <u>3.347</u>.

Titration is a commonplace laboratory technique of quantitative chemical analysis to determine the attention of an identified analyte. A reagent, termed the titrant or titrator, is ready as a trendy answer of recognized awareness and extent.

<u>Calculation:-</u>

Normality of acid                                               Normality of base

= nMV                                                                        nMV

= 1 × 0. 15 × 0.017                                              1 ×  0. 20 ×0.015 L

= 2.55 × 10⁻³                                                             = 3 × 10⁻³

The overall base will be high

net concentration = 3× 10⁻³ - 2.55 × 10⁻³

                             = 0.45 × 10⁻³

                             = 4.5× 10⁻⁴

pH = -log[4.5 × 10⁻⁴]

    = 4 - log4.4

     = <u>3.347</u>

A titration is defined as 'the manner of determining the amount of a substance A by using adding measured increments of substance B, the titrant, with which it reacts till precise chemical equivalence is completed the equivalence factor.

Learn more about titration here:-brainly.com/question/186765

#SPJ4

7 0
1 year ago
Which of the following is a weak base ?
Scrat [10]
<span>NH3 is the answer, if u look up its ph its a weak base</span>
7 0
3 years ago
Read 2 more answers
How many moles of glucose are in 19.1g of glucose?​
Crank

Answer:

0.106 mol (3s.f.)

Explanation:

To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.

Ar of C= 12, Ar of H= 1, Mr of O= 16

These Ar values can be found on the periodic table.

Mr of glucose= 6(12)+ 12(1) + 6(16)= 180

Moles of glucose

= mass ÷ mr

= 19.1 ÷ 180

= 0.106 mol (3 s.f.)

3 0
3 years ago
A mixture is made by combining 1.62 lb of salt and 5.20 lb of water.
WINSTONCH [101]

Answer:

23.8

Explanation:

Formula

weight % = weight of solute/ weight of solution x 100

weight of solution = weight of salt + weight of water

weight of solution = 1.62 lb + 5.20 lb = 6.82 lb

weight % = 1.62 / 6,82 x 100

weight % = 0.238 x 100

weight % = 23.8

4 0
3 years ago
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