Aluminum flouride+ sulfuric acid=

Brut [27]

Scientist rely on their estimates when they cannot obtain on exact numbers

7 0

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]

<u>Answer:</u>

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

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

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

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

<u>Explanation:</u>

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

<u>For a:</u>

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.

<u>For b:</u>

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.

<u>For c:</u>

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.

<u>For d:</u>

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.

7 0

3 years ago

A lab analysis of 0.502 g of Scandium (Sc) reaction with Oxygen (O2) from Air resulted

VARVARA [1.3K]

The empirical formula of the metal oxide would be $Sc_2O_3$

<h3>What is empirical formula?</h3>

Mass of scandium = 0.502g

Mass of scandium + oxygen = 0.768g

Mass of oxygen = 0.768 - 0.502 = 0.266g

Mole of scandium = 0.502/44.95 = 0.011

Mole of oxygen = 0.266/16 = 0.017

Divide by the lowest:

Scandium = 0.011/0.011 = 1

Oxygen = 0.017.0.011 = 1.5

Thus, the empirical formula of the compound would be $ScO_1.5$ or $Sc_2O_3$

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3 0

3 years ago

The effective nuclear charge for an atom is less than the actual nuclear charge due to

Irina18 [472]

Answer:

A = shielding.

Explanation:

The addition of electron causes the atomic size increase from top to bottom due to increase in atomic number.

As the atomic number increased one more electron is added and because of this electron on more electronic shell is added. Thus the electron become more away from the nucleus as many of other electrons are present in the way from nucleus to the outer electrons.

The hold of nucleus becomes weaker. Although nuclear charge is also increased but at the same time other electrons shield the respective electrons. So effective nuclear charge is weaker than the actual nuclear charge.

Because of this shielding it is easy to remove the electrons or we can say ionization energy decreases.

3 0

4 years ago

Does any solid Ag₂CrO₄ form when 2.7x10⁻⁵g of AgNO₃ is dissolved in 15.0 mL of 4.0x10⁻⁴MK₂CrO₄?

e-lub [12.9K]

Molarity of Ag+ is less than the molar solubility thus ppt will not occur.

Balanced reaction-:

Moles of AgNO3=mass(g)molar mass (g/mol) =2.7×10−5g / 169.86 gmol

=1.589⋅10^−7 mol

Molarity of Ag+=moles of solute(L)=1.589⋅10−7 mol0.015 L=1.059⋅10−5M

Ksp of Ag2CrO4

=[Ag+]2[CrO42−]

1.2⋅10−12=[2s]2[s]

4s3=1.2⋅10−12

s=6.69⋅10−5 M

Molarity of Ag+ is less than the molar solubility thus ppt will not occur.

<h3>What is the molarity calculation formula?</h3>

The volume of solvent required to dissolve the provided solute is multiplied by the ratio of the moles of the solute whose molarity has to be computed. (M=frac{n}{V}) The molality of the solution that needs to be computed in this case is M. n is the solute's molecular weight in moles.

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3 0

1 year ago