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nevsk [136]
3 years ago
7

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
 <span>The high-energy electron travels down an electron transport chain, losing energy as it goes.
Some of the released energy drives pumping of </span><span><span>\text H^+<span>H<span><span>​+</span><span>​​</span></span></span></span>H, start superscript, plus, end superscript</span><span> ions from the stroma into the thylakoid interior, building a gradient.
</span><span><span>H^+<span>H<span><span>​+</span><span>​​</span></span></span></span>H, start superscript, plus, end superscript</span><span> ions from the splitting of water also add to the gradient.
</span><span><span> H^+<span>H<span><span>​+</span><span>​​</span></span></span></span>H, start superscript, plus, end superscript</span><span> ions flow down their gradient and into the stroma, they pass through ATP synthase, driving ATP production in a process known as </span>chemiosmosis<span>.</span>
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Valuable ore deposits and gem crystals are often associated with _____. a. oceans b.oil deposits c. thin crustal areas d. igneou
elixir [45]
For the first question:
Valuable ore deposits and gem crystals are often associated with igneous intrusions. So for this question the correct option is option “d”. An example like the Pegmatite’s, which are a form of igneous intrusions, is responsible for the creation of variety of gems like topaz and tourmaline.
For the second question:
For this specific question, option “c” seems to be the correct option. Weathering, deposition, compaction, cementation are the possible sequence that leads to the formation of sedimentary rocks.





3 0
3 years ago
Read 2 more answers
What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V2O5 with 4.0 moles of
IRINA_888 [86]

Answer:

Theoretical yield of vanadium = 1.6 moles

Explanation:

Moles of V_2O_5 = 1.0 moles

Moles of Ca = 4.0 moles

According to the given reaction:-

V_2O_5_{(s)} + 5Ca_{(l)}\rightarrow 2V_{(l)} + 5CaO_{(s)}

1 mole of V_2O_5 react with 5 moles of Ca

Moles of Ca available = 4.0 moles

Limiting reagent is the one which is present in small amount. Thus, Ca is limiting reagent. (4.0 < 5)

The formation of the product is governed by the limiting reagent. So,

5 moles of Ca on reaction forms 2 moles of V

1 mole of Ca on reaction for 2/5 mole of V

4.0 mole of Ca on reaction for \frac{2}{5}\times 4 mole of V

Moles of V = 1.6 moles

<u>Theoretical yield of vanadium = 1.6 moles</u>

4 0
3 years ago
For which one of the following is ΔHfo zero?
Vlada [557]
The answer is most likely C
6 0
3 years ago
Read 2 more answers
II. Ionic Equations
mario62 [17]

Answer:

Complete ionic: \begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Net ionic: \begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

Explanation:

Start by identifying species that exist as ions. In general, such species include:

  • Soluble salts.
  • Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2. These three salts will exist as ions:

  • Each \rm AgNO_3\, (aq) formula unit will exist as one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion.
  • Each \rm CaCl_2 formula unit will exist as one \rm Ca^{2+} ion and two \rm Cl^{-} ions (note the subscript in the formula \rm CaCl_2\!.)
  • Each \rm Ca(NO_3)_2 formula unit will exist as one \rm Ca^{2+} and two \rm {NO_3}^{-} ions.

On the other hand, \rm AgCl is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2 (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each \rm AgNO_3\, (aq) formula unit will exist as only one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion. However, because the coefficient of \rm AgNO_3\, (aq)\! in the original equation is two, \!\rm AgNO_3\, (aq) alone should correspond to two \rm Ag^{+}\! ions and two \rm {NO_3}^{-}\! ions.

Do not rewrite the salt \rm AgCl because it is insoluble.

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of \rm Ca^{2+} and two units of \rm {NO_3}^{-}. Doing so will give:

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}.

Simplify the coefficients:

\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

7 0
2 years ago
A stock solution has a concentration of 1.5 M NaCl and is diluted to a 0.80 M solution with a volume of 0.10 L. What volume of t
ioda

Answer:

0.053 L  is the volume of concentrated solution that was used

Explanation:

Let's determine the answer of this, by rules of three.

There is also a dilution formula.

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution.

In 1 L of concentrated solution, there are 1.5 moles of NaCl

In 1 L of diluted solution, there are 0.80 moles.

The volume for the diluted solution is 0.10L

The rule of three will be:

1L of solution has 0.80 moles of solute

Then, 0.10L of solution must have (0.1 . 0.8)/1 = 0.08 moles

This moles came from the concentrated solution, and we know that in 1L of this solution we have 1.5 moles. Therefore the rule of three will be:

1.5 moles are in 1L of solution

0.08 moles were in (0.08 . 1L / 1.5) = 0.053 L (This is the volume of concentrated solution that was used)

Dilution formula is: M conc . Vol conc = M diluted . Vol diluted

1.5 M . Vol conc = 0.80 M . 0.10L

Vol conc = 0.80 M . 0.10L / 1.5M = 0.053L

4 0
3 years ago
Read 2 more answers
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