What is the uncertainty of 300.0ft

Enter your answer in the provided box. Oxidation of gaseous ClF by F2 yields liquid ClF3, an important fluorinating agent. Use t

MakcuM [25]

Answer : The value of $\Delta H_{rxn}$ for the reaction is, -135.2 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of $ClF_3$ will be,

$ClF(g)+F_2(g)\rightarrow ClF_3(l)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $2ClF(g)+O_2(g)\rightarrow Cl_2O(g)+OF_2(g)$ $\Delta H_1=167.5kJ$

(2) $2F_2(g)+O_2(g)\rightarrow 2OF_2(g)$ $\Delta H_2=-43.5kJ$

(3) $2ClF_3(l)+2O_2(g)\rightarrow Cl_2O(g)+3OF_2(g)$ $\Delta H_3=394.4kJ$

We are dividing the reaction 1, 2 and 3 and reversing reaction 3 and then adding all the equations, we get :

(1) $ClF(g)+\frac{1}{2}O_2(g)\rightarrow \frac{1}{2}Cl_2O(g)+\frac{1}{2}OF_2(g)$ $\Delta H_1=\frac{167.5kJ}{2}=83.75kJ$

(2) $F_2(g)+\frac{1}{2}O_2(g)\rightarrow OF_2(g)$ $\Delta H_2=\frac{-43.5kJ}{2}=-21.75kJ$

(3) $\frac{1}{2}Cl_2O(g)+\frac{3}{2}OF_2(g)\rightarrow ClF_3(l)+O_2(g)$ $\Delta H_3=\frac{-394.4kJ}{2}=-197.2kJ$

The expression for enthalpy of formation of $ClF_3$ will be,

$\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(83.75kJ)+(-21.75kJ)+(-197.2kJ)$

$\Delta H_{rxn}=-135.2kJ$

Therefore, the value of $\Delta H_{rxn}$ for the reaction is, -135.2 kJ

5 0

3 years ago

How do the energy and the most probable location of an electron in the third shell of an atom compare to the energy and the most

solmaris [256]

The answer is (2). You can think about this question in terms of the Bohr's model of the atom or in terms of quantum chemistry. In the Bohr model, electrons exist in discrete "shells," each respresenting a fixed spherical distance from the nucleus in which electrons of certain energy levels orbit the nucleus. The larger the shell (the greater the "orbit" radius), the greater the energy of the "orbiting" electron (I use quotations because electrons don't actually orbit the nucleus in the traditional sense, as you may know). Thus, according to the Bohr model, a third shell electron should be farther from the nucleus and have greater energy than an electron in the first shell. The quantum model is differs drastically from the Bohr model in many ways, but the essence is the same. A larger principal quantum number indicates 1) greater overall energy and 2) a probability distribution spread a bit more outward.

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3 years ago

Value of MA is always lesser thàn 1 why?

Degger [83]

Answer:

because the load arm is greater than the effort arm. As we know that when the load arm is greater than the effort arm, the mechanical advantage will always be lesser than one, which results in gain in speed

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pickupchik [31]

Active is more active which means it's in person

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3 years ago

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Q#5 Balance and list the coefficients from reactants to products.

skad [1K]

A. $2Fe_2O_3(s) +3C(s)$ → $4Fe(s) + 3CO_2(g)$