How to find the average speed of something

Follow the Carbon in Citric Acid Cycle: 1. How many cycles does it take to convert Acetyl CoA to CO2

Anastasy [175]

Each round of Citric Acid Cycle produce 2 molecules of carbon dioxide. So one cycle of TCA cycle is enough to convert Acetyl CoA to carbon dioxide.

<h3>Explanation: </h3>

Citric Acid Cycle or Kreb's Cycle or the TCA cycle is the 1st dedicated step towards the aerobic respiration. The end product of glycolysis is Pyruvate which is a three carbon compound. It's acted upon by Pyruvate Decarboxylase to produce a 2 carbon compound Acetyl CoA and a molecule of carbon dioxide. This Acetyl CoA now reacts with oxaloacetate to produce Citric Acid which is the 1st step of Citric Acid Cycle. This now produce several intermediates and a lot of reduced electron carriers along with 2 molecules of carbon dioxide and ends up being oxaloacetate again. So one cycle of Citric Acid Cycle is necessary to convert Acetyl CoA to CO2.

6 0

3 years ago

Calculate the number of grams of carbon dioxide produced from 6 moles of oxygen gas. List only the number of grams of carbon dio

faust18 [17]

Answer:

g CO2(g) = 264.06 g

Explanation:

C(s) + O2(g) → CO2(g)

∴ moles O2(g) = 6 mol

∴ mm CO2 = 44.01 g/mol

⇒ mol CO2(g) = ( 6 mol O2(g) )×( mol CO2(g) / mol O2(g) ) = 6 mol CO2(g)

⇒ mass CO2(g) = ( 6 mol CO2(g) )×( 44.01 g/mol ) = 264.06 g CO2(g)

3 0

3 years ago

How was your day????????

Trava [24]

Answer:

I literally just woke up for school hehe

3 0

3 years ago

Read 2 more answers

Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di

katrin [286]

Answer: The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

Explanation:

Calculating the molarity of solution:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

$\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M$

Calculating the molarity of solution:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

$0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g$

To calculate the molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (camphor) = 70 g

$M_{solute}$ = Molar mass of solute (camphor) = 152.2 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

$\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m$

Calculating the mole fraction of camphor:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For camphor:

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol$

For ethanol:

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

$\chi_{(camphor)}=\frac{0.459}{10.272}=0.045$ \

Calculating the mass percent of camphor:

To calculate the mass percentage of camphor in solution, we use the equation:

$\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100$

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

$\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%$

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

3 0

3 years ago

2Na3N3 -> 2Na + 3N2 if 0.58 mol if NaN3 reacts, what mass of nitrogen would result

Ronch [10]

Answer:

Mass = 24.36 g of N₂

Explanation:

The balance chemical equation for the decomposition of NaNO₃ is as follow;

2 NaN₃ → 2 Na + 3 N₂

Step 1: Find moles of N₂ as;

According to equation,

2 moles of NaNO₃ produces = 3 moles of N₂

So,

0.58 moles of NaNO₃ will produce = X moles of N₂

Solving for X,

X = 3 mol × 0.58 mol / 2 mol

X = 0.87 mol of N₂

Step 2: Calculate mass of N₂ as,

Mass = Moles × M.Mass

Mass = 0.87 mol × 28.01 g/mol

Mass = 24.36 g of N₂

8 0

3 years ago