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zysi [14]
2 years ago
11

The fat adipose cells is broken down into fatty acids. Which pathway is involved in this process?

Chemistry
1 answer:
VladimirAG [237]2 years ago
3 0

Answer:

the anwer is glycolysis

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Is slime a liquid or solid?
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Answer:

a <u>non-</u><u> </u>NEWTONIAN liquid, which means viscosity, and or resistance, of the liquid changes as you apply stronger force

8 0
3 years ago
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What step is ussaly done right before the experimentation in the scientific method
barxatty [35]

The Step that is Usually done right before the experimentation in the Scientific Method is Developing a Hypothesis and asking a Question.

Hope that helped!

~Izzy


3 0
3 years ago
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If 10.7 grams of NH4Cl is dissolved in enough water to make 800 mL of solution,what will be it's molarity
olga nikolaevna [1]
<h3><u>Answer;</u></h3>

Molarity = 0.25 M

<h3><u>Explanation;</u></h3>

Molarity is given by moles/Liter.

First we find moles:

Number of moles = Mass /molar mass

= (10.7g NH4Cl)/(53.5g/mol NH4Cl)

= 0.200 moles NH4Cl  

Then  we convert to liters:

= (800mL)*(1L/1000mL) = 0.800L  

Therefore; molarity = 0.2moles/0.8L

                                = 0.25M

6 0
3 years ago
Which particles may be gained lost or shared by an atom when it forms a chemical bond?
Maru [420]

Answer:

B.

Explanation:

electrons can be lost by one particle, and gained by another particle

6 0
3 years ago
he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
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