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ankoles [38]
1 year ago
14

After completion of the citric acid cycle, most of the usable energy from the original glucose molecule is in the form of ______

____.
A. acetyl CoA
B. ATP
C. CO2
D. FADH2
E. NADH
Chemistry
1 answer:
GaryK [48]1 year ago
5 0

After completion of the citric acid cycle, most of the usable energy from the original glucose molecule is in the form of B. ATP

The most basic kind of sugar, monosaccharides only have one kind of sugar molecule. The simplest sugar is glucose, which is also your body's primary energy source. In tests to determine blood sugar levels, sugar is measured. Fructose and galactose, two more monosaccharides, are converted into glucose via metabolism ( 1 , 2 ). For instance, ATP is necessary for both breathing and keeping your heart beating. As well as assisting in the synthesis of lipids and nerve impulses, ATP also facilitates the entry and exit of certain molecules from cells. Even some living things, including bioluminescent creatures like jellyfish and fireflies, use ATP to create light!

Learn more abut ATP here:

brainly.com/question/14637256

#SPJ4

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I have no idea how to do this and need help
den301095 [7]

Answer:

60.55%

Explanation:

nO3=155/48

nO2 used: 3/2nO3=4.84375

percent yield: 4.84375/8=60.55%

5 0
3 years ago
Explain, in terms of electrons, why the bonding in strontium sulfide, SrS, is similar to the bonding in magnesium bromide, MgBr2
Aleksandr-060686 [28]

Answer: both compounds have ionic bond between metal and non-metal

Explanation: both Sr and Mg are earth alkaline metals and form ions Mg^2+

And Sr^2+. Br forms ion Br^- and S ion is S^2+.

5 0
3 years ago
The The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k=0.029 W/ m measured temperature differ
vfiekz [6]

Answer:

Heat flux = 13.92 W/m2

Rate of heat transfer throug the 3m x 3m sheet = 125.28 W

The thermal resistance of the 3x3m sheet is 0.0958 K/W

Explanation:

The rate of heat transfer through a 3m x 3m sheet of insulation can be calculated as:

q=-k*A*\frac{\Delta T}{\Delta X}\\\\q=-0.029\frac{W}{m*K}*(3m*3m)*\frac{12K}{0.025m}  =125.28W

The heat flux can be defined as the amount of heat flow by unit of area.

Using the previous calculation, we can estimate the heat flux:

heat \, flux=\frac{q}{A}=\frac{125.28 W}{9 m^{2} }  =13.92 W/m^{2}

It can also be calculated as:

q/A=-k*\frac{\Delta T}{\Delta X}

The thermal resistance can be expressed as

\Delta T=R_t*Q\\R_t=\Delta T/Q=\frac{\Delta X}{k*A}

For the 3m x 3m sheet, the thermal resistance is

R_t = \frac{\Delta X}{k*A}=\frac{0.025m}{0.029W/mK*9m^{2}}=0.0958 \, K/W

4 0
3 years ago
Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio
umka21 [38]

Answer : The value of K_p at this temperature is 66.7

Explanation : Given,

Pressure of PCl_3 at equilibrium = 0.348 atm

Pressure of Cl_2 at equilibrium = 0.441 atm

Pressure of PCl_5 at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}

Now put all the values in this expression, we get :

K_p=\frac{(10.24)}{(0.348)(0.441)}

K_p=66.7

Therefore, the value of K_p at this temperature is 66.7

4 0
3 years ago
How much pv work is done in kilojoules for the reaction of 0.68 mol of h2 with 0.34 mol of o2 at atmospheric pressure if the vol
dsp73
The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much

<span>actual delta H = 0.34  moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>

<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>
3 0
3 years ago
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