HELP!!!!!!!!!! PLEASE!!! HELP!!

The light reactions of photosynthesis use _____ and produce _____. The light reactions of photosynthesis use _____ and produce _

SashulF [63]

<h2>Input = $NADP^+$ , water and Output = $NADPH + O_2$ </h2>

Explanation:

The light reactions of photosynthesis use water and produce Oxygen, NADPH.

The equation for photosynthesis :

$6 CO_2 + 6 H_2O$ → $C_6H_12O_6 + 6 O_2$

The process of photosynthesis in two stages -

The first stage is called the light reaction in which the light energy from the sun is captured and converted into chemical energy stored in the form of ATP and NADPH

The second stage is the process of conversion of ATP molecules to sugar or glucose (the Calvin Cycle)

For a light reaction -

Net Input is of, $NADP^+$ , $Light, Water, ADP$

Net Output is of, $ATP, NADPH, O_2$

8 0

3 years ago

The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

Na: 23
H: 1
C: 12
O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So the limiting reagent is sodium bicarbonate.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

So the grams of excess reactant that do not participate in the reaction are 0333 g.

3 0

3 years ago

How many moles of silver are there in 6.9 x 10^28 silver atoms?

Juli2301 [7.4K]

Answer:

1.2 x 10⁵ moles Ag (2 sig. figs.)

Explanation:

1 mole any substance (elements or compounds) => 6.023 x 10²³ particles of specified substance

∴ 6.9 x 10²⁸ atoms Ag = 6.9 x 10²⁸ Ag atoms / 6.023 x 10²³ Ag atoms/mole Ag

= 1.145608501 x 10⁵ moles Ag (calculator answer)

= 1.2 x 10⁵ moles Ag (2 sig. figs.)

6 0

2 years ago

Read 2 more answers

Nuclear Processes Unit Test

Finger [1]

Answer:

?

Explanation:

8 0

3 years ago

Read 2 more answers

What is one of the main features of karst topography?

Sauron [17]

Answer:

Karst is a topography formed from the dissolution of soluble rocks such as limestone, dolomite, and gypsum. It is characterized by underground drainage systems with sinkholes and caves. It has also been documented for more weathering-resistant rocks, such as quartzite, given the right conditions.

4 0

3 years ago

Read 2 more answers