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3 years ago

What is the word eletrovent

Chemistry

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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Calculate the mass of vanadium(V) oxide (V2O5) that contains a million (1.0 *10^6) vanadium atoms. Be sure your answer has a uni

sweet [91]

Answer : The mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

Explanation : Given,

Number of atoms of $V_2O_5$ = $1.0\times 10^{6}$

Molar mass of $V_2O_5$ = 181.88 g/mole

In $V_2O_5$ , there are 2 atoms of vanadium and 5 atoms of oxygen.

First we have to determine the moles of $V_2O_5$ .

As, $2\times 6.022\times 10^{23}$ number of vanadium atom present in 1 moles of $V_2O_5$

So, $1.0\times 10^{6}$ number of vanadium atom present in $\frac{1.0\times 10^{6}}{2\times 6.022\times 10^{23}}=8.3\times 10^{-19}$ moles of $V_2O_5$

Now we have to determine the mass of $V_2O_5$ .

$\text{Mass of }V_2O_5=\text{Moles of }V_2O_5\times \text{Molar mass of }V_2O_5$

$\text{Mass of }V_2O_5=(8.3\times 10^{-19}mole)\times (181.88g/mole)=1.51\times 10^{-16}g$

Therefore, the mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

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4 years ago

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As frequency of waves increases, wavelength __________. Question 21 options: decreases increases becomes faster remains constant

tensa zangetsu [6.8K]

Frequency decreases when wavelength increase

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3 years ago

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Though alchemists were often superstitious, they left a rich legacy for modern chemists. What was their main contribution?

topjm [15]

though alchemists were often superstitious, they left a rich legacy of modern chemists. what was their main contribution-

Explanation:

they were the first to preform experiments.

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3 years ago

Give the formulas of the compounds in each set (a) lead(l|) oxide and lead(lV) oxide; (b) lithium nitride, lithium nitrite, and

Ivanshal [37]

Answer:

For a: The formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b: The formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c: The formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d: The formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

Explanation:

All the given compounds are ionic compounds. This means that between the atoms complete sharing of electrons takes place.

For a:

Lead is the 82th element of periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^2$ .

To form $Pb^{2+}$ ion, this element will loose 2 electrons and to form $Pb^{4+}$ ion, this element will loose 4 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lead (II) oxide is PbO and for lead (IV) oxide is $PbO_2$

Thus, the formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b:

Lithium is the 3rd element of periodic table having electronic configuration of $[He]2s^1$ .

To form $Li^{+}$ ion, this element will loose 1 electron.

Nitrogen is the 7th element of periodic table having electronic configuration of $[He]2s^22p^3$ .

To form $N^{3-}$ ion, this element will gain 3 electrons.

Nitrite ion is a polyatomic ion having chemical formula of $NO_2^{-}$

Nitrate ion is a polyatomic ion having chemical formula of $NO_3^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lithium nitride is $Li_3N$ , for lithium nitrite is $LiNO_2$ and for lithium nitrate is $LiNO_3$

Thus, the formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c:

Strontium is the 38th element of periodic table having electronic configuration of $[Kr]5s^2$ .

To form $Sr^{2+}$ ion, this element will loose 2 electrons.

Hydrogen is the 1st element of periodic table having electronic configuration of $1s^1$ .

To form $H^{-}$ ion, this element will gain 1 electron and is named as hydride ion.

Hydroxide ion is a polyatomic ion having chemical formula of $OH^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for strontium hydride is $SrH_2$ and for strontium hydroxide is $Sr(OH)_2$

Thus, the formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d:

Magnesium is the 12th element of periodic table having electronic configuration of $[Ne]3s^2$ .

To form $Mg^{2+}$ ion, this element will loose 2 electrons.

Manganese is the 25th element of the periodic table having electronic configuration of $[Ar]3d^54s^2$

To form $Mn^{2+}$ ion, this element will loose 2 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for magnesium oxide is MgO and for manganese (II) oxide is MnO.

Thus, the formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

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3 years ago

1) The real basis of periodicity of elements is the electron

SIZIF [17.4K]

Answer:

charge

Explanation:

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