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Vera_Pavlovna [14]

4 years ago

8

For the reaction 2HNO3 + Mg(OH)2 → Mg(NO3)2 + 2H2O, how many grams of magnesium nitrate are produced from 18.21 mol of nitric ac

id, HNO3?
a. 1350
b. 901
c. 5410
d. 1130

1 answer:

enyata [817]4 years ago

7 0

The answer is A. 1350

(ALL NUMBERS Except the grams is given from the equation)
18.21mol HNO3/1 * 1mol MgN206/2mol/ 2mol HNO3 * (molar mass-->)148.30 MgN206/ 1mol MgN2O6

You multiply 18.21HNO3* 1mol MgN2O6 * 148.30MgN2O6
Then divide it by the 2mol HNO3 to get 1350

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3 years ago

Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+

DIA [1.3K]

These are four questons and four answers:

Answers:

1) 7⁺

2) 1⁻

3) 3⁺

4) 5⁺

Explanation:

<u><em>Question 1) </em></u><u><em>Cl₂O₇:</em></u>

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

2×X + 7 (-2) = 0

2X - 14 = 0

2X = +14

X = +14 /2 = + 7

<em>Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.</em>

<u><em>Question 2) </em></u><u><em>AlCl₄⁻</em></u>

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

1 (+3) + 4X = - 1

+3 + 4X = - 1

4X = - 1 - 3

4X = - 4

X = - 1

<em>Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.</em>

<em><u>Question 3)</u></em><em><u> Ba(ClO₂)₂</u></em>

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

+2 + 2X + 4 (-2) = 0

2X +2 - 8 = 0

2X - 6 = 0

2X = +6

X = + 3

<em>Conclusion: the oxidation number of Cl in Ba(ClO₂)₂ is 3⁺.</em>

<u><em>Question 4)</em></u><u><em> CIF₄⁺</em></u>

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

X + 4(-1) = +1

X - 4 = +1

X = +1 + 4

X = + 5

<em>Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.</em>

6 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