Câu 15: Nguyên tử X nặng gấp 2 lần nguyên tử cacbon. Nguyên tử X đó có nguyên tử khối và

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

4 years ago

What type of reaction is CuO + H2SO4 ➡️ CuSO4 + H2O

Afina-wow [57]

Water is the answer hope this helps

7 0

3 years ago

A 1.2516 gram sample of a mixture of caco3 and na2so4 was analyzed by dissolving the sample and completely precipitating the ca

Dennis_Churaev [7]

Answer:

0.009725 moles of H2C2O4

0.009725 moles CaCO3

Mass percentage = 77.77%

Explanation:

Step 1: The balanced equation

2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O

We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O

Step 2: Calculate moles of MnO4-

Molarity = Moles/volume

Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-

Moles of Mno4- = 0.1092M * 35.62 *10^-3 L

Moles of MnO4- = 0.00389 moles

Step 3: Calculate moles of H2C2O4

Since there is needed 5 moles of C2H2O4 to consume 2 moles of MnO4-

then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = 0.009725 moles of H2C2O4

Step 4: Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

Step 5: Calculate mass of CaCO3

Molar mass of CaCO3 = 100.09 g/mole

Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3

Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g

Step 6: Calculate percentage by weight of CaCO3

Mass of CaCO3 = 0.9734g

Mass of original sample = 1.2516g

Mass percentage = 0.9734/1.2516 *100% = 77.77%

6 0

4 years ago

What is the name of an ideal-gas process in which no heat is transferred?.

PIT_PIT [208]

Answer:

Adiabatic. This is a process where no heat is being added or removed from the system. Or can be simply stated as: no heat transfer (or heat flow) happening in a system.

Explanation:

6 0

2 years ago

Please help ASAP. No one could not help me with this question.

Tanya [424]

Answer:

sorry

Explanation:

I don't know the answer this is really confusing but I am really sorry you have to do this.

3 0

3 years ago