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MrMuchimi
3 years ago
7

A sample of ethanol (C2H6O) has a mass of

Chemistry
2 answers:
kupik [55]3 years ago
7 0

Answer:

6.52 kj

Explanation:

Just got it correct on edge 2020 :)

Tems11 [23]3 years ago
3 0

Answer:

6.52 kJ

Explanation:

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Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th
Black_prince [1.1K]

Answer:

K_{goal}=1.793*10^{-33}

Explanation:

N2(g)+O2(g)⇌2NO(g), K_1 = 4.10*10^{-31}

N2(g)+2H2(g)⇌N2H4(g), K_2 = 7.40*10^{-26}

2H2O(g)⇌2H2(g)+O2(g), K_3 = 1.06*10^{-10}

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g)                     Eq (1)

Equilibrium constant for Eq (1) is K_1*K_3*K_3

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g)                       Eq (2)  

Equilibrium constant for Eq (2) is (K_1*K_3*K_3)^{1/2}

Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}

7 0
3 years ago
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

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Hydrogen i suppose is the right one
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3 years ago
Công thức của A có dạng Ca(hco3)x có ptk là 162 tìm x
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Answer:

j

Explanation:

nwjwjw

dodkekds

dldldle

5 0
2 years ago
C.) Is heating of cobalt (ii) chloride crystals a chemical change or a physical change (Imk)​
miv72 [106K]

Answer: Heating the hydrated forms of cobalt chloride reverses the reactions above, returning cobalt chloride to the blue, water-free, or anhydrous, state. Water is "liberated" in these reactions, known as dehydration reactions.

Explanation:

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