How many atp molecules are produced during the citric acid cycle?

Chemistry

2 answers:

Lesechka [4]2 years ago

8 0

Answer:

2 on edge2020

Explanation:

Mrrafil [7]2 years ago

7 0

"Only 2 molecules" of ATP produced during the citric acid cycle

Hope this helps!

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The critical point of carbon dioxide is 304 k and 73 atm. under which conditions is carbon dioxide a liquid? i. 303 k and 73 atm

Sergeeva-Olga [200]

Answer is: a) I only.
Above critical temperature of CO₂, a gas cannot be liquefied no matter how much pressure is applied. Temperature and pressure above its critical point is called supercritical fluid and this is intermediate between gaseous and liquid states.

7 0

2 years ago

How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2

Nuetrik [128]

Answer:

Moles of $ScCl_3$ = 6 moles

Explanation:

The reaction of $Sc$ and $Cl_2$ to make $ScCl_3$ is:

$2Sc+3Cl_2$ ⇒ $2ScCl_3$

The above reaction shows that 2 moles of Sc can react with 3 moles of $Cl_2$ to form $ScCl_3.$

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of $Cl_2$ = $Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ = $10 *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ =15 moles

So 15 moles of $Cl_2$ are required to react with 10 moles of $Sc$ but we have 9 moles of $Cl_2$ , it means $Cl_2$ is limiting reactant.

$Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *\frac{2\ Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}$

$Moles\ of\ ScCl_3=9 *\frac{2\ Moles\ of\ ScCl_3}{3\ Moles\ of\ Cl_2}$

Moles of ScCl_3= 6 moles

4 0

2 years ago

Read 2 more answers

Using only the 1h nmr spectrum of the crude nitration product, determine if the product is mostly ortho-, meta-, or para- substi

tatiyna

The nitartion of methyl benzoate is expected to proceed as given in the equation below:

In methyl benzoate there are 3 types of 1 H proton. The two ortho to the C=O group is a doublet at 8 ppm the 2 metal to the C=O is a multiple at 7.5 ppm and one para to the C=O is a multiplet at 7.5 ppm.

On nitration the ortho will probably show two signal one being a single with 3 proton integration and one a doublet with 1 H integration

The meta will show a highly down field singlet (coresponding to 1 proton), two unequal doublets (corresponding to 1 H each) and one multiplets (corresponding to 1H). This is the major product as seen from the 1H NMR.

The para isomer will come as two doublets which will be very close to each other there is a small signal for this set between 8.2 and 8.3 ppm.