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tatuchka [14]
3 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]3 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]3 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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Which shows the formula for an organic acid?
Inga [223]

Answer:

A

Explanation:

CH3CH2COOH- Propanoic Acid

6 0
3 years ago
Why is important to only test one variable at a time?
Natalija [7]

Answer:

B

Explanation:

It is important to only test one variable at a time because you need to be able to disprove or prove a problem with just one independent variable. When you have several variables in the experiment, it would be impossible to know which variable honestly caused the end result.

4 0
2 years ago
in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

3 0
3 years ago
How many molecules units are in 4.5 moles of CF4
kobusy [5.1K]
1.0 mole ---------- 6.02x10²³ molecules
4.5 moles -------- ?

4.5 * 6,02x10²³ / 1.0

= 2.709x10²⁴ molecules units
7 0
3 years ago
Given the following chemical equation, if 50.1 grams of silicon dioxide is heated with excess carbon and 32.3 grams of silicon c
aivan3 [116]

Answer:

97%.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

SiO2 (s) + 3C (s) —> SiC(s) + 2CO(g)

Next, we shall determine the mass of SO2 that reacted and the mass of SiC produced from the balanced equation. This is illustrated below:

Molar mass of SiO2 = 28 + (16x2) = 60 g/mol

Mass of SO2 from the balanced equation = 1 x 60 = 60 g

Molar mass of SiC = 28 + 12 = 40 g/mol

Mass of SiC from the balanced equation = 1 x 40 = 40 g.

From the balanced equation above,

60 g of SiO2 reacted to produce 40 g of SiC.

Next, we shall determine the theoretical yield of SiC. This can be obtained as follow:

From the balanced equation above,

60 g of SiO2 reacted to produce 40 g of SiC.

Therefore, 50.1 g of SiO2 will react to produce = (50.1 x 40)/60 = 33.4 g of SiC.

Therefore, the theoretical yield of SiC is 33.4 g

Finally, we shall determine the percentage yield of SiC as follow:

Actual yield of SiC = 32.3 g

Theoretical yield of SiC = 33.4 g

Percentage yield =?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 32.3/33.4 x 100

Percentage yield = 96.7 ≈ 97%

Therefore, the percentage yield of the reaction is 97%.

3 0
3 years ago
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