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3 years ago

Which process in the light-dependent reactions results in the release of hydrogen ions, electrons, and oxygen?

Chemistry

2 answers:

MariettaO [177]3 years ago

5 0

The correct answer is :

electron transport

Explanation:

The light-dependent reactions utilize light energy to make two molecules required for the next stage of photosynthesis: the energy accommodation molecule ATP and the diminished electron carrier NADPH. In plants, the light reactions take place in the thylakoid membranes of organelles termed as chloroplasts.

Electron carriers are decreased during glycolysis and the citric acid cycle to NADH + H+ and FADH2. Certain carriers then donate electrons and protons to the atom carrier proteins of the electron transport chain. The last electron acceptor is oxygen. Coincidentally with oxygen, electrons and protons form molecules of water.

vodomira [7]3 years ago

3 0

The high-energy electron travels down an electron transport chain, losing energy as it goes.
Some of the released energy drives pumping of \text H^+H+H, start superscript, plus, end superscript ions from the stroma into the thylakoid interior, building a gradient.
H^+H+H, start superscript, plus, end superscript ions from the splitting of water also add to the gradient.
 H^+H+H, start superscript, plus, end superscript ions flow down their gradient and into the stroma, they pass through ATP synthase, driving ATP production in a process known as chemiosmosis.

You might be interested in

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

The mass of $H_2O$ is,

$m_{H_{2} O}$ = 6.31 mol $\times$ 18.02g/ mol

= 114g

On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

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8 0

2 years ago

Please help me to answer this question. i have to submit tomorrow.

ratelena [41]

Answer:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

4 0

3 years ago

Aspirin tablets contain the active ingredient (aspirin) and various inactive ingredients. Suggest what types of inactive ingredi

Debora [2.8K]

Binders ; fillers. They are inert substances to hold pill shape and preserve the medication, its ph / etc.

5 0

3 years ago

What occurs during one half-life?

lorasvet [3.4K]

The half-life of a radioactive compound is the time taken for that said isotope to decay or disintegrate so that only half of the initial atoms remain in that compound. During the decay process, the isotope will give off energy and matter, and the way to depict this is indicated by t 1/2.

8 0

4 years ago

Read 2 more answers

A rock contains 0.623 mg of 206Pb for every 1.000 mg of 238U present. Assuming that no lead was originally present, that all the

maxonik [38]

Answer:

t = 3,496x10⁹ years

Explanation:

The decay of ²³⁸U is:

²³⁸U → ²⁰⁶Pb + 8He + 6e⁻

Moles of ²⁰⁶Pb presents in 0,623mg are:

0,623x10⁻³g×(1mol / 206g) = 3,02x10⁻⁶ moles of ²⁰⁶Pb.

These moles are equals to moles of ²³⁸U before decay, that means, 3,02x10⁻⁶ moles²³⁸U

In grams:

3,02x10⁻⁶ moles²³⁸U× (238g / 1mol) = 7,20x10⁻⁴ g ²³⁸U = 0,720 mg²³⁸U

That means initial ²³⁸U was 1,000mg + 0,720mg = 1,720mg

Applying the formula:

ln (N₀/N) t₁₂ = t ln2

Where N₀ is initial amount of uranium (1,720mg), N is concentration of uranium (1,000mg), half-life time is a constant (t₁₂= 4,468x10⁹ years) and t is the time transcurred for the reaction. Replacing:

ln(1,720/1)*4,468x10⁹ years = t ln2

t = 3,496x10⁹ years

I hope it helps!

6 0

3 years ago