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3 years ago

How many turns of the citric acid cycle must you complete?

2 answers:

8 0

Two turns are needed because glycolysis produces two pyruvic acid molecules when it splits glucose.

____ [38]3 years ago

7 0

Two turns are needed because glycolysis produces two pyruvic acid molecules

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What mass, in grams, of carbon dioxide gas can be produced by complete combustion of 24.0 grams of carbon?

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3 years ago

Please select the word from the list that best fits the definition

AveGali [126]

Answer:

I choose D option because may be it's correct

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3 years ago

Read 2 more answers

The question is below

rodikova [14]

The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

Equations:

Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V

The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

3 0

3 years ago

Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0

3 years ago

What did Kwang Jeon discover when researching amoeba infected by bacteria?

defon

Kwang Jeon observed that Amoeba had been attacked by a bacterial infection, and lots of the Amoeba had died. However, some survived and continued to reproduce. After investigating the remaining Amoeba and their offspring, he noticed they were very healthy. He thought maybe they were able to fight off the bacteria, but instead, he found they were still infected with the bacteria but were not dying. The bacteria were no longer making the Amoeba sick. Then, he killed off the bacteria using antibiotics and was surprised to see that the Amoeba also died. It seemed the Amoeba and bacteria had formed a relationship in which they both needed each other to survive. After researching, Jeon found that the bacteria made a protein that the Amoeba needed to survive.

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3 years ago