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Anna71 [15]
3 years ago
8

2. Why should a magnesium ribbon be cleaned before burning in air ?​

Chemistry
1 answer:
Nesterboy [21]3 years ago
8 0
To remove magnesium oxide layer from the ribbon which may prevent or slow down the burning of magnesium ribbon.
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Using the equations 2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/m
Dafna11 [192]

Answer : The enthalpy for the reaction is 49.1 kJ/mol

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 reaction of C_6H_6 will be,

6C(s)+3H_2(g)\rightarrow C_6H_6(l)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) 2C_6H_6(g)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)    

\Delta H_1=-6271kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    

\Delta H_2=-393.5kJ/mole

(3) 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)    

\Delta H_3=-483.6kJ/mole

Now we will reverse the reaction 1 and divide by 2, multiply reaction 2 by 6 and reaction 3 by 3/2 then adding all the equations, we get :

(1) 6CO_2(g)+3H_2O(l)\rightarrow C_6H_6(g)+\frac{15}{2}O_2(g)    

\Delta H_1=-\frac{-6271kJ/mole}{2}=3135.5kJ/mol

(2) 6C(s)+6O_2(g)\rightarrow 6CO_2(g)    

\Delta H_2=6\times (-393.5kJ/mole)=-2361kJ/mol

(3) 3H_2(g)+\frac{3}{2}O_2(g)\rightarrow 3H_2O(l)    

\Delta H_3=\farc{3}{2}\times (-483.6kJ/mole)=-725.4kJ/mol

The expression for enthalpy of formation of C_6H_6 will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(3135.5kJ/mole)+(-2361kJ/mole)+(-725.4kJ/mole)

\Delta H=49.1kJ/mole

Therefore, the enthalpy for the reaction is 49.1 kJ/mol

7 0
3 years ago
Which are formed as glaciers retreat and leave debris behind in mounds at the front of their farthest advance?
Natali5045456 [20]
It seems that you have missed the given options for this question which are the following: <span>A. glacial striations B. terminal moraines C. eskers D. cirques
Anyway, the correct answer would be option B. The one that are formed as glaciers retreat and leave debris behind in mounds at the front of their farthest advance are terminal moraines. Hope this answer helps.</span>
7 0
3 years ago
The materials created from a chemical reaction are called
sweet [91]
The answer would be products
8 0
2 years ago
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When a sample of rust was mixed with acid, the rust changed into two simpler substances: W and Z. The substances W and Z could n
julia-pushkina [17]

here can be a number of chemical processes by which a compounds can be broken down into simpler substances. The most common and wide-spread of such processes is Catabolism. Along with anabolism, catabolism make up the metabolism process for living organisms. In catabolism, complex chemical molecules (such as proteins, polysaccharides, etc.) are broken down into simpler molecules (such as amino acids, monosaccharides, etc.). This is often accompanied by release of energy in the form of ATP (adenosine triphosphate) molecules and intermediate metabolites (which can be used by the organism in the anabolic processes). The energy thus generated is used for operation and maintenance of cells (and consequently, the body).  

Other processes that break down chemical substances into simpler substances, include depolymerization, decomposition, etc.

Hope this helps.

6 0
3 years ago
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For this discussion, respond to the following... An electron falls through a distance d in a uniform electric field of magnitude
mars1129 [50]

The electron should experience a greater acceleration due to it's significantly smaller mass and should fall through distance "d" in a shorter amount of time.

<u>Explanation:</u>

The electron force can be expressed as F=qE. According to Newton's second law of motion force can be expressed as F=ma. This can be written as a=F/m. Substituting electric force expression for "F" in this equation, we get a=qE/m. This means acceleration is conversely proportional to mass and directly to electric field and charge. This means that proton having significantly larger mass than electron should experience smaller amount of acceleration and would take longer to fall at distance "d".

On the other hand, the electron would experience greater acceleration due to it's significantly smaller mass and would fall faster at distance "d", unlike the situation of proton.

4 0
2 years ago
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