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3 years ago

Into which molecule are all the carbon atoms in glucose ultimately incorporated during cellular respiration?

Chemistry

1 answer:

pav-90 [236]3 years ago

3 0

Answer:

The correct option is: carbon dioxide (CO₂)

Explanation:

Cellular respiration refers to the process of generation of energy in the form of adenosine triphosphate (ATP) from various nutrients such as carbohydrates, fats, and proteins.

Aerobic cellular respiration involves the oxidation of nutrients such as glucose (C₆H₁₂O₆), by molecular oxygen (O₂) to give carbon dioxide (CO₂), water (H₂O) and heat energy.

Reaction involved: (glucose) C₆H₁₂O₆ (s) + 6 O₂ (g) → 6 CO₂ (g) + 6 H₂O (l) + heat energy

Therefore, in a cellular respiration, all the carbon atoms of the glucose molecules, form carbon dioxide.

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A scientist wants to make a solution of tribasic sodium phosphate, na3po4, for a laboratory experiment. How many grams of na3po4

natima [27]

Answer :

The correct answer is for mass of Na₃PO₄ 39.7 g.

Given : 1) Molarity of Na⁺ ions = 1.00 M or 1.00 $\frac{mol}{L}$

2) Volume of solution = 725 mL

Converting volume of solution from mL to L :

Conversion factor : 1 L = 1000 mL

$Volume of solution = 725 mL * \frac{1L }{1000 mL}$

Volume of solution = 0.725 L

Following steps can be done to find mass of :

Step 1 : Write the dissociation reaction of Na₃PO₄ .

$Na_3PO_4 3 Na^+ + PO_4^3^-$

Step 2: Find moles of Na⁺ ions :

Mole of Na⁺ ions can be calculated using molarity formula which is : $Molarity (\frac{mol}{L} ) = \frac{mole of Na^+ }{volume of solution }$

Plugging value of Molarity and volume

$1.00 \frac{mol}{L} = \frac{Mole of Na^+ ions}{0.725 L}$

Multiplying both side by 0.725 L

$1.00 \frac{mol}{L}* 0.725 L = \frac{mole of Na^+ ions}{0.725 L} * 0.725 L$

Mole of Na⁺ ions = 0.725 mol

Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :

Mole ratio is found from coefficients from balanced reaction as:

Mole of Na₃PO₄ in balanced reaction = 1

Mole of Na⁺ ion = 3

Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3

Step 4 : To find mole of Na₃PO₄

Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :

$Mole of Na_3PO_4= Mole of Na^+ * Mole ratio$

$Mole of Na_3PO_4 = 0.725 mol * \frac{1 mole of Na_3PO_4}{3 mole of Na^+ }$

Mole of Na₃PO₄ = 0.242 mol

Step 5 : To find mass of Na₃PO₄

Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using molar mass of Na₃PO₄ as :

$Mass (g) = mole (mol) * molar mass \frac{g}{mol}$

$Mass of Na_3PO_4 = 0.242 mol * 163.94 \frac{g}{mol}$

Mass of Na₃PO₄ = 39.619 g

Step 6 : To round off mass of Na₃PO₄ to correct sig fig .

The sig fig in 750 mL and 1.00 M is 3 . So mass of Na₃PO₄ can rounded to 3 sig fig as :

Mass of Na₃PO₄ = 39.7 g

8 0

2 years ago

1 mole of no2(g) has a greater entropy than 1 mole of n2o4(g). true or false

bija089 [108]

In order to answer this, you need to find the empirical data for the standard entropies. Please refer to this link: http://www.mrbigler.com/misc/energy-of-formation.PDF

For NO₂ gas, the entropy is 240 J/mol-K. For N₂O₄ gas, the entropy is 304.2 J/mol-K. Therefore, the statement is false.

7 0

2 years ago

Nitric acid (HNO3) is a strong acid, and it is titrated with a standard solution of sodium hydroxide (NaOH), a strong base. Whic

aev [14]

C. pH = 7.0

i just had the test

4 0

3 years ago

Read 2 more answers

What particle limitation to the widespread of this reusable water bottles? Pls help someone

Sauron [17]

Answer: Plastic water bottles

Explanation:

If you use disposable water bottles, here are some important concerns you should know about how they’re made as well as the problems they cause for the planet, your health, and your wallet.

3 0

3 years ago

A 100.0 ml sample of 0.10 m ca(oh)2 is titrated with 0.10 m hbr. determine the ph of the solution after the addition of 400.0 ml

WINSTONCH [101]

The balanced equation for the reaction is as follows;
Ca(OH)₂ + 2HBr --> CaBr₂ + 2H₂O
stoichiometry of Ca(OH)₂ to HBr is 1:2
number of Ca(OH)₂ moles reacted - 0.10 mol/L x 0.1000 L = 0.010 mol
Number of HBr moles added - 0.10 mol/L x 0.4000 = 0.040 mol
1 mol of Ca(OH)₂ needs 2 mol of HBr for neutralisation
therefore 0.010 mol of Ca(OH)₂ needs - 0.010 x 2 = 0.020 mol of HBr to be neutralised
but 0.040 mol of HBr has been added therefore number of moles of HBr in excess - 0.040 - 0.020 = 0.020 mol
then pH of the medium can be calculated using the excess H⁺ ions
HBr is a strong acid therefore complete ionization
[HBr] = [H⁺]
[H⁺] = 0.020 mol / (100.0 + 400.0 mL)
= 0.020 mol / 0.5 L
= 0.040 mol/L
pH = -log[H⁺]
pH = - log [0.040 M]
pH = 1.40
pH of the medium is 1.40

8 0

3 years ago