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tatuchka [14]
4 years ago
7

For each molecule of glucose processed during glycolysis, the net yield is ____. 1. two molecules of NADH, two of ATP, and two o

f pyruvate2. two molecules of NAD , two of ATP, and two of pyruvate 3. two molecules of NAD , four of ATP, and two of pyruvate 4. two molecules of NADH, four of ATP, and two of pyruvate
Chemistry
2 answers:
Lana71 [14]4 years ago
5 0

Answer: 1. two molecules of NADH, two of ATP, and two of pyruvate

Explanation:

During glycolysis 4 molecules of ATP are synthesized by two substrate level phosphorylation (first in step 6 conversion of 1,3-Bisphospho glycerate to 3-Bisphospho glycerate by 1,3-Bisphospho glycerate kinase produces 2 molecules of ATP . Secondly 2 molecules of ATP is produced in step 9 by pyruvate kinase). But 2 molecules of ATP are used in step 1 and 3 by hexokinase and phospho fructo kinase respectively, hence a net yield of only two ATP.

2 molecules of NADH is produced in step 5 by Glyceraldehyde-3-phosphate dehydrogenase.

The end product of glycolysis is 2 molecules of pyruvate.

Natali5045456 [20]4 years ago
5 0

Answer:

1. two molecules of NADH, two of ATP, and two of pyruvate.

Explanation:

Glycolysis is defined as the breakdown of sugar which takes place in the cytoplasm. The various processes and enzymes involved are summarised in the Kreb's cycle.

Glycolysis is present in nearly all living organisms. Glucose is the source of almost all energy used by cells. Overall, glycolysis produces two pyruvate molecules, a net gain of two ATP molecules, and two NADH molecules.

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Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati
ollegr [7]

Explanation:

Let us assume that the given data is as follows.

   mass of barium acetate = 2.19 g

   volume = 150 ml = 0.150 L    (as 1 L = 1000 ml)

   concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

        Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}

Hence, moles of C_{2}H_{3}O^{-}_{2} = 2 \times Ba(C_{2}H_{3}O_{2})_{2}  .......... (1)

As,   No. of moles = \frac{mass}{\text{molar mass}}

Hence, moles of Ba(C_{2}H_{3}O_{2})_{2} will be calculated as follows.                          

     No. of moles = \frac{mass}{\text{molar mass}}  

                          =  \frac{2.19 g}{255.415 g/mol}   (molar mass of Ba(C_{2}H_{3}O_{2})_{2} is 255.415 g/mol)            

                       = 8.57 \times 10^{-3}

    Moles of C_{2}H_{3}O^{-}_{2} = 2 \times 8.57 \times 10^{-3}

                          = 0.01715 mol

Hence, final molarity will be as follows.

              Molarity = \frac{\text{no. of moles}}{volume}

                             = \frac{0.01715 mol}{0.150 L}

                             = 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

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3 years ago
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