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3 years ago

13

What is the word eletrovent

1 answer:

djyliett [7]3 years ago

8 0

Answer:

Explanation:

Electrovalent is a word often associated with chemical bonding in the field of chemistry. It is special type of bond that occurs between metals and non-metals.

These bond types are interatomic interactions occurring between two atoms to ensure that they attain stable configurations.

This bond type is also known as ionic bonds.
It occurs between two species with a large electronegative diffference i.e one specie is electropositive and the other highly electronegative.
The more electropositive specie is metal and it readily loses its valence electrons.
The electronegative non-metal gains the electrons and becomes negatively charged.
The electrostatic attraction between the metal and non-metals yields the electrovalent bonds.

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The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane wi

Setler79 [48]

Answer:

Explanation:

mass fraction N₂ : He : CH₄ : C₂H₆ : : 15 : 5 : 60 : 20

mole fraction N₂ : He : CH₄ : C₂H₆ : : 15/28 : 5/4 : 60/16 : 20/30

mole fraction N₂ : He : CH₄ : C₂H₆ : : .5357 : 1.25 : 3.75 : .67

Total mole fractions = .5357 + 1.25 + 3.75 + 0.67 = 6.2057

mole fraction of N₂ = .5357 / 6.2057 = .0877

mole fraction of He = 1.25 / 6.2057 = .20

mole fraction of CH₄ = 3.75 / 6.2057 = .6043

mole fraction of C₂H₆ = .67 / 6.2057 = .108

Partial pressure = total pressure x mole fraction

Partial pressure of N₂ = 1200 kPa x .0877 = 105.24 kPa

Partial pressure of He = 1200 kPa x .20 = 240 kPa

Partial pressure of CH₄ = 1200 kPa x .6043 = 725.16 kPa

Partial pressure of C₂H₆ = 1200 kPa x .108 = 129.6 kPa

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3 years ago

In the following reaction, how many grams of oxygen will react with 10.47 grams of benzene (C6H6)? 2C6H6 + 1502 12CO2 + 6H2O The

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4 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

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4 years ago