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Fynjy0 [20]
3 years ago
9

Xenon (xe) and fluorine (f) can combine to form several different compounds, including xef2, which contains 0.2894 g fluorine fo

r every gram of xenon. use the law of multiple proportions to determine, n, which represents the number of f atoms in another compound, xefn, given that it contains 0.8682 g of f for every gram of xe.
Chemistry
2 answers:
mihalych1998 [28]3 years ago
8 0

We can say that for every 2 atoms of F, there is a mass of 0.2894 g as evident from the compound XeF2. Now to find for f, we can say that there are n atoms of F for every 0.8682 g of F. Taking the proportion:

 

2 atoms / 0.2894 g = n / 0.8682 g

 

Calculating for n:

n = 2 * 0.8682 / 0.2894

n = 6

oee [108]3 years ago
7 0

Answer is: n (number of fluorine atoms) is 6, formula of the compound is XeF₆.

m(F) = 0.8682 g; mass of fluorine.

n(F) = m(F) ÷ M(F).

n(F) = 0.8682 g ÷ 19 g/mol.

n(F) = 0.0457 mol; amount of substance.

m(Xe) = 1.00 g.

n(Xe) = 1.00 g ÷ 131.293 g/mol.

n(Xe) = 0.00761 mol.

n(Xe) : n(F) = 0.00761 mol : 0.0457 mol.

n(Xe) : n(F) = 1 mol : 6 mol.

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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} }}

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

Let say the assumed temperature = 25° C

= (25+ 273)K

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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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Answer:

The final temperature is 31.95° C.

Explanation:

Given that,

Initial temperature of a sample of chloroform, T_i=25^{\circ} C

Mass of chloroform, m = 150 g

It absorbs 1 kJ of heat, Q = 10³ J

The specific heat of chloroform, c = 00.96 J/gºC

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Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\\dfrac{Q}{mc}+T_i=T_f\\\\T_f=\dfrac{10^3}{150\times 0.96}+25\\\\=31.95^{\circ} C

So, the final temperature is 31.95° C.

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