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2 years ago

What are the products of the light reactions?

2 answers:

max2010maxim [7]2 years ago

8 0

During these initial reactions, water is used and oxygen is released. The energy from sunlight is converted into a small amount of ATP and an energy carrier called NADPH. Together with carbon dioxide, these are used to make glucose (sugar) through a process called the Calvin Cycle.

Anna [14]2 years ago

4 0

Answer:

ATP, NADPH and oxygen are the products of light reaction.

Explanation:

Light reaction occurs in the presence of sunlight and is the first phase of photosynthesis reaction. Chlorophyll molecule absorbs the energy from the sunlight and later uses in the formation of ATP and NADPH.

The water is used up during the reaction. The light energy is used to obtain ATP, NADPH and oxygen. These products of light reaction are used further by plants during dark reaction of photosynthesis.

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NaOH+HCl->............

Oxana [17]

Answer:

NaCl+H20

Explanation:

It is a neutralisation reaction in which NaOH is a base and HCl is an acid. On reaction it forms salt and water.

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3 0

2 years ago

Read 2 more answers

BARSIC [14]

The number of C atoms in 0.524 moles of C is 3.15 atoms.

The number of $SO_2$ molecules in 9.87 moles $SO_2$ is 59.43 molecules.

The moles of Fe in 1.40 x $10^{22}$ atoms of Fe is 0.23 x $10^{-1}$

The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ is 3.81.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × $10^{23}$ x 0.524 mole

3.155601758 atoms =3.155 atoms

B. The number of $SO_2$ molecules in 9.87 moles of $SO_2$ :

6.02214076 × $10^{23}$ x 9.87

59.4385293 molecules= 59.43 molecules

C. The moles of Fe in 1.40 x $10^{22}$ atoms of Fe:

1.40 x $10^{22}$ ÷ 6.02214076 × $10^{23}$

0.2324754694 x $10^{-1}$ moles.

0.23 x $10^{-1}$ moles.

D. The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ :

2.30x $10^{24}$ ÷ 6.02214076 × $10^{23}$

3.819239854 moles=3.81 moles

Learn more about moles here:

brainly.com/question/8455949

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6 0

1 year ago

A solution has a pH of 11.8. What is the pOH?

amid [387]

Answer:

pOH is 2.2

Explanation:

The pH scale goes from 0 to 14. As such, the pH and the pOH add up to 14. Subtract your pH from 14 to get your answer.

14 - 11.8 = 2.2

5 0

2 years ago

If a nanometer is one billionth of a meter (0.000 000 00 1 m), how many nanometers are there in one meter?

Anit [1.1K]

Answer:

One billion

Explanation:

A nanometer is a BILLIONTH of a meter which means it takes one billion to equal a meter

6 0

3 years ago

If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t

tankabanditka [31]

Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

$Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3$

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of $CaCO_3$

$\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}$

Given:

Mass of $CaCO_3$ = 0.524 g

Molar mass of $CaCO_3$ = 100 g/mol

$\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol$

Now we have to calculate the concentration of $CaCO_3$

$\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M$

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

$CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}$

So,

Concentration of calcium ion = Concentration of $CaCO_3$ = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

4 0

3 years ago