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2 years ago

Which equation represents a neutralization reaction ?

Chemistry

2 answers:

Tpy6a [65]2 years ago

8 0

Answer:

C) Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2H₂O

Explanation:

Neutralization reaction:

When an acid and base react they form water and salt.The reaction is called neutralization reaction.

General equation:

HX + BOH → BX + H₂O

From given reaction equations only option C is correct because only this reaction gives salt and water.

Chemical equation:

Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2H₂O

Calcium hydroxide is acting as a base. Sulfuric acid is an acid. Both these reactants react to form salt which is calcium sulfate and water.

Calcium hydroxide dissociate to gives Ca²⁺ ion and OH⁻ while H₂SO₄ dissociate to gives H⁺ and SO₄²⁻. Ca²⁺ and SO₄²⁻ react to form salt CaSO₄ while OH⁻ and H⁺ form water.

iren [92.7K]2 years ago

7 0

Answer:Ba(OH)2+Cu(NO3)2—>Ba(NO3)2+Cu(OH)2

Explanation:

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Fe3+(aq)+6H2O(l)⇌Fe(H2O)63+(aq) : F e 3 + ( a q ) + 6 H 2 O ( l ) ⇌ F e ( H 2 O ) 6 3 + ( a q ) : blank is the Lewis acid and bl

Tasya [4]

Answer:

Lewis acid- Fe3+

Lewis base- water molecule

Explanation:

Acids and bases have been defined in diverse ways. There have been definitions put forward by Arrhenius, Brownstead and Lowry as well as Lewis. Each definition his useful in its own way.

Lewis acids are lone pair acceptors such as metal ions. This implies that in the particular instance of this reaction, Fe3+ is the lewis acid.

Similarly, a Lewis base is a lone pair donor, all ligands are lone pair donors since they donate one or more lone pairs of electrons to Lewis acids. In the particular instance of this reaction, the Lewis base is the water molecule.

6 0

2 years ago

Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu+ ion), (b) manganese oxide (containi

ivann1987 [24]

Answer:The formulas of ionic compounds are:

a) $CuBr$

b) $Mn_2O_3$

c) $Hg_2I_2$

d) $Mg_3(PO_4)_2$

Explanation:

Formulas for the an ionic compounds is determine 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.

(a) Copper bromide :Given that it contains $Cu^+$ ion.

$Cu^++Br^-\rightarrow CuBr$

(b) Manganese oxide : Given that it contains $Mn^{3+}$ ion.

$Mn^{3+}+O^{2-}\rightarrow Mn_2O_3$

(c)Mercury iodide :Given that it contains $Hg_2^{2+}$

$Hg_2^{2+}+I^-\rightarrow Hg_2I_2$

(d) Magnesium phosphate :Given that it contains $PO_4^{3-}$

$Mg^{2+}+PO_4^{3-}\rightarrow Mg_3(PO_4)_2$

4 0

2 years ago

Sodium oxide (Na2O) crystallizes in a structure in which the O2– ions are in a face - centered cubic lattice and the Na+ ions ar

melisa1 [442]

We have to know the number of Na⁺ ions in the unit cell.

The number of Na⁺ ions in the unit cell is (D) 8.

Sodium oxide (Na2O) crystallizes in a structure in which the O2– ions are in a face - centered cubic lattice and the Na+ ions are in tetrahedral holes.

O²⁻ ions are in a face centred cubic lattice, so the number of O²⁻ ions per unit cell is equal to 4. The number of tetrahedral hole= 2 X 4=8. Na+ ions are present in tetrahedral holes, which indicates there are 8 number of Na+ ions in the unit cell.

3 0

2 years ago

When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6

Mariulka [41]

Answer:

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

Explanation:

Wavelength of the photon emitted = $\lambda =646.3 nm =646.3\times 10^{-9} m$

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

$E=\frac{hc}{\lambda }$

h = Planck's constant = $6.626\tiomes 10^{-34} Js$

c = Speed of the light = $3\times 10^8 m/s$

$E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}$

$E=3.07\times 10^{-19} J$

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

3 0

2 years ago

A precipitate of zinc hydroxide can be formed using the reaction below.

worty [1.4K]

Answer:

Option B is correct. KOH is the limiting reagent, and 0.27 mole of Zn(OH)2 precipitate is produced.

Explanation:

Step 1: Data given

Number of moles ZnCl2 = 0.36 moles

Number of moles KOH = 0.54 moles

Step 2: The balanced equations

ZnCl2(aq) + 2 KOH(aq) → Zn(OH)2(s) + 2 KCl(aq)

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

Step 3: Calculate the limiting reactant.

KOH is the limiting reactant. It will completely be consumed (0.54 moles). ZnCl2 is in excess. There will react 0.54/2 = 0.27 moles

There will remain 0.36 - 0.27 = 0.09 moles.

Step 4: The products

There will be produced 2 moles KCl and 1 mol Zn(OH)2. Zn(OH)2 is the precipitate produced.

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

For 0.54 moles KOH, we will produce 0.27 moles precipitate (Zn(OH)2)

Option A is not correct because 0.27 mol of Zn(OH)2 will precipitate, not 0.54 mol

Option B is correct

Option C is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

Option D is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

7 0

2 years ago