# Which of the following four statements is/are accurate with respect to glycolysis? A. Glycolysis involves the conversion of monosaccharides into glucose. B. Glycolysis involves the breakdown of glucose into pyruvate. C. Glycolysis involves the conversion of pyruvate into glucose to glyceraldehyde-3-phosphate. D. Glycolysis involves the conversion of pyruvate into acetyl CoA.

B. Glycolysis involves the breakdown of glucose into pyruvate.

Explanation:

Glycolysis is the biological process whereby one glucose molecule is broken down to two pyruvate molecules. There are a lot of enzymatic processes that are involved in this. It is one of the most important reactions in the world because it allows living cells to harness chemical energy from organic molecules

## Related Questions

How many molecules are in 0.400 moles of. N2O5?

2.40 × 10²³ Molecules of N₂O₅

Explanation:

As we know 1 mole of any substance contains 6.022 × 10²³ particles (also called as Avogadro's Number). Therefore, the relation between moles and number of particles can be written as,

Moles  =  Number of Particles ÷ 6.022 ×10²³ Particles.mol⁻¹

For the molecules of Dinitrogen pentoxide (N₂O₅) it can be written as,

Moles  =  Number of Molecules ÷ 6.022 ×10²³ Molecules.mol⁻¹

Rearranging for Number of Molecules,

Number of Molecules  =  Moles × 6.022 ×10²³ Molecules.mol⁻¹

Putting value of Moles,

Number of Molecules  =  0.400 mol × 6.022 ×10²³ Molecules.mol⁻¹

Number of Molecules  =  2.40 × 10²³ Molecules of N₂O₅

Because you're turning moles into molecules, all you need to use is Avogadro's Number (6.022*10^23).
The equation would be: .400 mol N2O5/1 mol*6.022*10^23 molecules.
The answer, including significant figures, is 151 molecules.

Suppose you use 0.0150 liter of a 2.50 M solution of silver nitrate. Assuming the reaction goes completion how much silver sulfide is produced

C. 9.30 g

M= Moles/ L

2.50=x/0.0150

x=0.0375 moles

moles x mm= grams

0.0375 x 247.8= 9.2925 grams

for the reaction shown compute the theoretical yield of product in moles each of the initial quantities of reactants. 2 Mn(s)+3 O2 (g) _____ MnO2(s) 2 mol Mn , 2 mol O2

Answer: The theoretical yield of manganese dioxide is 57.42 grams

Explanation:

We are given:

Moles of manganese = 2 moles

Moles of oxygen gas = 2 moles

For the given chemical equation:

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 2 moles of manganese

So, 2 moles of oxygen gas will react with = of manganese

As, given amount of manganese is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 1 mole of manganese dioxide

So, 2 moles of oxygen gas will produce = of manganese dioxide

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

Molar mass of manganese dioxide = 87 g/mol

Moles of manganese dioxide = 0.66 moles

Putting values in above equation, we get:

Hence, the theoretical yield of manganese dioxide is 57.42 grams

2 mole MnO₂

Explanation:

2Mn(s) + 2O₂(g) => 2MnO₂(s)

Question 21 0.1 pts
Given the following equation:
2 Al(s) +3 FeO(s) 3Fe(s)+ Al2O3(s)
How many moles of aluminum are needed to react completely with 1.2 moles of
FeO?
24n
1.2 n
oooo
8n
1.6n