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Vlad1618 [11]
2 years ago
15

After passing through pyruvate dehydrogenase and the citric acid cycle, one mole of pyruvate will result in the formation of ___

_____ moles of carbon dioxide and ________ mole(s) of ATP (or GTP).
Chemistry
1 answer:
mamaluj [8]2 years ago
7 0

Answer:

The answer to be filled in the respective blanks in question is

3 and 1

Explanation:

So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.

Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1

 

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A sample of an unknown substance has a mass of .158 kg if 2,510 J of heat is required to heat the substance from 32°C to 61°C wh
ziro4ka [17]

Answer: 0.548J/g°C

Explanation:

Q = s × m × DeltaT

Q = Heat (J)

S = Specific Heat Capacity

M = mass (g)

DeltaT = Change in temperature (°C)

0.158Kg x 1000 = 158g

2.510J = s x 158g x (61°C-32°C)

2.510J/(158g x 29°C) = s

S = 0.54779.... J/g°C

S = 0.548 J/g°C

7 0
2 years ago
A gas occupies 22.4 l at stp and 14.5 l at 100c and 2.00 atm pressure. how many moles of gas did the system gain or lose?
azamat
<span>At standard temperature and pressure 22.4 l of an ideal gas would contain 1 mole. in order to find the change in moles we must look at the ideal gas law PV=nRT where P=Pressure V=volume n=Moles R= Gas constant T= Temperature. To simplify this equation we will be using the gas constant at .08206 L-atm/mol-K. We must first convert 100c to k which is 373.15. Then we can plug the values into our equation which gives us (2atm)(14.5 l)=(n)(.08206 L-atm/mol-K)(373.15). After some basic algebra we get the moles to equal roughly .95 which is .05 moles less than our original system.</span>
6 0
2 years ago
What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c
brilliants [131]

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and  (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

8 0
2 years ago
Read 2 more answers
0.01 M HCl solution has a pH of 2. Suppose that during the experiment, both the universal pH indicator and the cabbage indicator
NNADVOKAT [17]
It matches the universal pH indicator and is indicating the proper pH
5 0
2 years ago
Read 4 more answers
Select the correct set of quantum numbers (n, l, ml, ms) for the first electron removed in the formation of a cation for stronti
matrenka [14]

Answer:

5,0,0,-1/2

Explanation:

The quantum numbers are a way to characterize the electrons, and so, identify the region that it's more probable to find it (orbital). They are:

- Principal quantum number (n): represents the shell or level, and varies from 1 to 7, and are represented by the letter K, L, M, N, O, P, and Q.

- Azimuthal quantum number (l): represents the subshell or sublevel, and is represented by 0,1,2,3.., and for the letters s, p, d, f,...

- Magnetic quantum number (ml): represents the orbital. It varies from -l to +l passing by 0. Each orbital can have 2 electrons.

- Spin quantum number (ms): represents the spin of the electron. It can be +1/2 or -1/2.

The strontium has an atomic number equal to 38, by the Linus Pauling's diagram, the electronic distribution is:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²

The valence electron is at the subshell 5s, which has only one magnetic quantum number: 0. Because it has 2 electrons, the first one has spin =1/2, and the other -1/2. So the first electron of the formation of cation has quantum numbers:

n = 5; l = 0; ml = 0; ms = -1/2

7 0
2 years ago
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