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stepladder [879]
2 years ago
15

Refer to the reaction between magnesium and oxygen In the introduction. What elements is oxidizing?write the oxidation number fo

r the reactants and product to show the oxidation
Chemistry
1 answer:
erica [24]2 years ago
7 0

Answer:

Oxidizing element: Mg

Explanation:

Reaction of magnesium and oxygen can be written as:

2Mg +O_2 \rightarrow 2MgO

while writting chemical reaction between two atom or element then write their natural form i.e. how that atom or element exist in the nature.

eg. reaction of any element with Oxygen atom then we can not write directly O because oxygen exist in the nature in form of O_2

finding oxidation number each side:

Reactant side:

Mg=0;

O=0;

product side:

Mg=+2;

O=-2;

from above:

oxidation number of Mg is increasing hence this is oxidizing element

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Answer:

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Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x  -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0
3 years ago
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6 0
2 years ago
The base-dissociation constant, kb, for pyridine, c5h5n, is 1.4x10-9 the acid-dissociation constant, ka, for the pyridinium ion,
sp2606 [1]
We are given the base dissociation constant, Kb, for Pyridine (C5H5N) which is 1.4x10^-9. The acid dissociation constant, Ka for the Pyridium ion or the conjugate acid of Pyridine is to be determined. We know from our chemistry classes that:

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where Kw is always equal to 1x10^-14

so, to solve for Ka of Pyridium ion, substitute Kb to the equation together with Kw and solve for Ka:

1x10^-14 = 1.4x10^-9 * Ka
solve for Ka

Ka = 7.14x10^-6 

Therefore, the acid dissociation constant of Pyridinium ion is 7.14x10^-6.
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