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VMariaS [17]
3 years ago
15

why metal in the middle of metal activity series can't be obtained by heating ors in air please help me I will give u branist.fo

llow u back and give u like plz help me with a good answer​
Chemistry
1 answer:
Dmitrij [34]3 years ago
6 0

\huge\mathsf{\red{\underline{\underline{Answer}}}}

{\green{\dashrightarrow}}Metals such as iron, zinc, lead, copper, etc., are in the middle of the reactivity series. These are moderately reactive metals and are usually present as sulphides or carbonates. A metal is obtained from its ore by the processes of reduction or by electrolysis.

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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
A carbon compound with a covalently bonded chlorine or bromine is called _____. an amide an halocarbon an alcohol an aldehyde
german
Halogens are elements that can be found in group 7 of the periodic table. They have 7 electrons in their outer shell and thus can form only a single covalent bond with other elements. Examples of halogens include chlorine, bromine and fluorine. A carbon compound that is covalently bonded with chlorine or bromine is called a halocarbon.
7 0
3 years ago
What does covalent and ionic bond have to do with chemical bond?
krok68 [10]
Covalent and ionic bonds are two different types of chemical bonding.  Covalent bonds involve the sharing of electrons between 2 atoms while ionic bonds involve the complete transferring of electrons from one atom to another.  Covalent bonds usually form between two nonmetals while ionic bonds usually form between a metal and a nonmetal.   
I hope this helps.  Let me know if anything is unclear.  
5 0
2 years ago
A material has a volume of 63.0 cm3 and a mass of 28 grams.  What is the density of the material in g/cm3 to the correct number
dimaraw [331]

Answer:

0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).

Explanation:

  • Knowing that:

<em>d = m/V,</em>

where, d is the density of the material (g/cm³).

m is the mass of the material (m = 28 g).

V is the volume of the material (V = 63.0 cm³).

<em>∴ d = m/V </em>= (28 g)/(63.0 cm³) = <em>0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).</em>

7 0
2 years ago
The following equation is one way to prepare oxygen in a lab. 2KClO3 → 2KCl + 3O2 Molar mass Info: MM O2 = 32 g/mol MM KCl = 74
Alex73 [517]
From  the  equation  above   the  reacting   ratio  of  KClO3   to  O2  is  2:3 therefore  the  number  of  moles  of  oxygen  produced  is  ( 4 x3)/2 =  6 moles  since   four  moles  of  KClO3  was  consumed
mass=relative  formula mass  x  number  of  moles
That  is   32g/mol x 6  moles  =192grams


8 0
3 years ago
Read 2 more answers
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