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Yuri [45]
3 years ago
12

How many molecules of glucose are produced by each cycle of the light reaction?

Chemistry
1 answer:
disa [49]3 years ago
4 0
Zero (0) molecules of glucose are produced.
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It is used to measure size or distance.

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How are fireworks made ?
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4 0
3 years ago
Sodium hydroxide (a base) reacts with hydrochloric acid. After the reaction is complete, a pH indicator is added to the solution
fomenos

Answer:

hydrochloric acid

Explanation:

The hydrochloric acid in this reaction is the limiting reactant. A limiting reactant is the reactant that is used up in a chemical reaction. It determines the extent of the reaction.

Since the solution indicates a basic one after the end of the reaction, this suggests that more of the sodium hydroxide is still left unreacted with.

The reactant in excess supply here is the sodium hydroxide and the bulk of it is till left in solution.

7 0
3 years ago
How many moles of carbon atoms are present<br> in 5.4 moles of glucose?<br> Answer in units of mol.
Delvig [45]

Answer:

32.4 mol

Explanation:

Given data:

Number of moles of C atom present = ?

Number of moles of glucose = 5.4 mol

Solution:

Glucose formula = C₆H₁₂O₆

There are 6 moles of C atoms are present in one mole of glucose.

In 5.4 moles of glucose:

5.4 mol × 6 = 32.4 mol

3 0
3 years ago
Calculate the number of moles and the mass of the solute in each of the following solutions:
frosja888 [35]

<u>Answer:</u>

<u>For a:</u> The number of moles of KI are 2.7\times 10^{-5} and mass is 4.482\times 10^{-3}g

<u>For b:</u> The number of moles of sulfuric acid are 1.65\times 10^{-5} and mass is 1.617\times 10^{-3}g

<u>For c:</u> The number of moles of potassium chromate are 2.84\times 10^{-2} and mass is 5.51 g.

<u>For d:</u> The number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}    .....(1)

To calculate the number of moles of a substance, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(2)

  • <u>For a:</u>

Molarity of KI = 8.23\times 10^{-5}M

Volume of solution = 325 mL = 0.325 L     (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

8.25\times 10^{-5}mol/L=\frac{\text{Moles of KI}}{0.325L}\\\\\text{Moles of KI}=2.7\times 10^{-5}mol

Now, using equation 2, we get:

Moles of KI = 2.7\times 10^{-5}mol

Molar mass of KI = 166 g/mol

Putting values in equation 2, we get:

2.7\times 10^{-5}mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=4.482\times 10^{-3}g

Hence, the number of moles of KI are 2.7\times 10^{-5} and mass is 4.482\times 10^{-3}g

  • <u>For b:</u>

Molarity of sulfuric acid = 22\times 10^{-5}M

Volume of solution = 75 mL = 0.075 L

Putting values in equation 1, we get:

22\times 10^{-5}mol/L=\frac{\text{Moles of sulfuric acid}}{0.075L}\\\\\text{Moles of }H_2SO_4=1.65\times 10^{-5}mol

Now, using equation 2, we get:

Moles of sulfuric acid = 1.65\times 10^{-5}mol

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 2, we get:

1.65\times 10^{-5}mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=1.617\times 10^{-3}g

Hence, the number of moles of sulfuric acid are 1.65\times 10^{-5} and mass is 1.617\times 10^{-3}g

  • <u>For c:</u>

Molarity of potassium chromate = 0.1135M

Volume of solution = 0.250 L

Putting values in equation 1, we get:

0.1135mol/L=\frac{\text{Moles of }K_2CrO_4}{0.250L}\\\\\text{Moles of }K_2CrO_4=2.84\times 10^{-2}mol

Now, using equation 2, we get:

Moles of potassium chromate = 2.84\times 10^{-2}mol

Molar mass of potassium chromate = 194.2 g/mol

Putting values in equation 2, we get:

2.84\times 10^{-2}mol=\frac{\text{Mass of }K_2CrO_4}{194.2g/mol}\\\\\text{Mass of }K_2CrO_4=5.51g

Hence, the number of moles of potassium chromate are 2.84\times 10^{-2} and mass is 5.51 g.

  • <u>For d:</u>

Molarity of ammonium sulfate = 3.716 M

Volume of solution = 10.5 L

Putting values in equation 1, we get:

3.716mol/L=\frac{\text{Moles of }(NH_4)_2SO_4}{10.5L}\\\\\text{Moles of }(NH_4)_2SO_4=39.018mol

Now, using equation 2, we get:

Moles of ammonium sulfate = 39.018 mol

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 2, we get:

39.018mol=\frac{\text{Mass of }(NH_4)_2SO_4}{132.14g/mol}\\\\\text{Mass of }(NH_4)_2SO_4=5155.84g

Hence, the number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

3 0
4 years ago
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