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melisa1 [442]
2 years ago
11

What can you say about the reactants of photosynthesis and the products of cellular respiration?​

Chemistry
1 answer:
xz_007 [3.2K]2 years ago
6 0

Answer:

Photosynthesis converts carbon dioxide and water into oxygen and glucose. Glucose is used as food by the plant and oxygen is a by-product. Cellular respiration converts oxygen and glucose into water and carbon dioxide. Water and carbon dioxide are by-products and ATP is energy that is transformed from the process.

Explanation:

You might be interested in
What volume of hydrogen will be produced at STP by the reaction of 67.3 g of tin with excess water according to the following re
VikaD [51]

Answer:

Sn + 2H2O ==> Sn(OH)2 + 2H2

67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters

Explanation:

Sn + 2H2O ==> Sn(OH)2 + 2H2

67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters

5 0
3 years ago
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
Anna71 [15]

Answer:

31.652g of Na3PO4

Explanation:

We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:

Na3PO4 will dessicate in solution as follow:

Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.

Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e

xM Na3PO4 = (1.10 x 1)/3

xM Na3PO4 = 0.367M

Therefore, the molarity of Na3PO4 is 0.367M.

Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:

Molarity of Na3PO4 = 0.367M

Volume = 525mL = 525/1000 = 0.525L

Mole of Na3PO4 =..?

Molarity = mole /Volume

0.367 = mole /0.525

Cross multiply

Mole of Na3PO4 = 0.367 x 0.525

Mole of Na3PO4 = 0.193 mole.

Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:

Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol

Mole of Na3PO4 = 0.193 mole

Mass of Na3PO4 =.?

Mass = mole x molar mass

Mass of Na3PO4 = 0.193 x 164

Mass of Na3PO4 = 31.652g

Therefore, 31.652g of Na3PO4 is needed to prepare the solution.

6 0
3 years ago
Suppose NAD is unavailable because NADH cannot be oxidized due to a mutation in the NADH dehydrogenase (Complex I). If FAD could
Dovator [93]

Answer:

FAD substitution will produce 28 ATP instead of 36.

Explanation:

NAD and FAD are coenzymes involved in reversible oxidation and reduction reactions. These compounds are also known as electron carriers. However NADH produce 3 electrons in electron transport chain and FADH2 produce 2 electron beacuase it transfer the electrons to second complex in ETC.

Normal prduction of ATP from glucose;

2 cytoplasmic NADH formed in glycolysis         Each yields 2 ATP   +4

2 NADH formed in the oxidation of pyruvate Each yields 3 ATP         +6

2 FADH2 formed in the citric acid cycle         Each yields 2 ATP         +4

6 NADH formed in the citric acid cycle             Each yields 3 ATP         +18

2 ATP from glycolysis                                                                                   +2  

2 ATP from citric acid cycle                                                                          +2                            

                                                                    Net yield ATP +36

C6H12O6 + 6 CO2 + 36 ADP + 36 Pi ⇒6 CO2 + 6 H2O + 36 ATP

If we replace the NAD with FAD the total ATP production would be.

2 cytoplasmic FADH2 formed in glycolysis          Each yields 2 ATP            +4

2 FADH2 formed in the oxidation of pyruvate   Each yields 3 ATP    +4

2 FADH2 formed in the citric acid cycle            Each yields 2 ATP    +4

6 FADH2 formed in the citric acid cycle                  Each yields 2 ATP   +12

2 ATP from glycolysis                                                                                    

+2

2 ATP from citric acid cycle                                                                            +2

                                                                  <u>Net yield ATP +28</u>

C6H12O6 + 6 CO2 + 28ADP + 28 Pi ⇒6 CO2 + 6 H2O + 28 ATP

7 0
3 years ago
What is the purpose of the lab? Weather patterns
Sedaia [141]

Answer:

Here:

Explanation:

To familiarize students with experimental apparatus, the scientific method, and methods of data analysis so that they will have some idea of the inductive process by which the ideas were originated. To teach how to make careful experimental observations and how to think about and draw conclusions from such data.

7 0
3 years ago
A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the
Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, C_6H_4N_2O_4

Explanation :

First we have to calculate the mass of benzene.

\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}

\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g  = 0.04395 kg

Formula used :  

\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}

where,

\Delta T_f = change in freezing point  = 5.53-1.37=4.16^oC

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of benzene

Molal-freezing-point-depression constant (K_f) for benzene = 5.12^oC/m

m = molality

Now put all the given values in this formula, we get

4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}

\text{Molar mass of unknown compound}=180.6g/mol

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.575}{1.193}=2.99\approx 3

For H = \frac{2.4}{1.193}=2.01\approx 2

For N = \frac{1.193}{1.193}=1

For O = \frac{2.381}{1.193}=1.99\approx 2

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_3H_2N_1O_2

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{180.6}{84}=2

Molecular formula = (C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4

Therefore, the molecular of the compound is, C_6H_4N_2O_4

3 0
3 years ago
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