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Greeley [361]
3 years ago
14

Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F

alse" to indicate whether or not the stated reaction (or "no reaction") correctly corresponds to the expected observation in each case. Lead(II) nitrate and sodium chloride; No reaction occurs. Sodium bromide and hydrochloric acid; No reaction occurs. Nickel(II) chloride and lead(II) nitrate; Pb2+(aq) + 2Cl-(aq) --> PbCl2(s) Magnesium chloride and sodium hydroxide; No reaction occurs. Ammonium sulfate and barium nitrate; Ba2+(aq) + SO42-(aq) --> BaSO4(s)
Chemistry
1 answer:
nirvana33 [79]3 years ago
3 0

<u>Answer:</u>

<u>For 1:</u> The correct answer is False.

<u>For 2:</u> The correct answer is True.

<u>For 3:</u> The correct answer is True.

<u>For 4:</u> The correct answer is False.

<u>For 5:</u> The correct answer is True.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.  If no net ionic equation is formed, it is said that no reaction has occurred.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.

  • <u>For 1:</u> Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is False.

  • <u>For 2:</u> Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)

Ionic form of the above equation follows:

Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

  • <u>For 3:</u> Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+Ni^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^{2+}(aq.)+2NO_3^-(aq.)

As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is True.

  • <u>For 4:</u> Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Mg^{2+}(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)

Hence, the correct answer is False.

  • <u>For 5:</u> Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)

Ionic form of the above equation follows:

2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)

Hence, the correct answer is True.

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1 . a chemical bond formed by the electrostatic attraction between ions covalent bond
liberstina [14]

Explanation:

When there occurs sharing of electrons between two chemically combining atoms then it forms a covalent bond. Generally, a covalent bond is formed between two non-metals.

An ionic bond is defined as the bond formed due to transfer of one or more number of electrons from one atom to another. An ionic bond is always formed between a metal and a non-metal.

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A molecule is defined as the smallest particle present in a substance or atom.

A metallic bond is formed due to mobile valence electrons shared by positive nuclei in a metallic crystal.

Thus, we can conclude that given statements are correctly matched as follows.

1).  a chemical bond formed by the electrostatic attraction between ions - ionic bond

2).  a chemical bond formed by two electrons that are shared between two atoms - covalent bond

3). the orbitals of an atom where electrons are found - energy level

4).  the smallest particle of a covalently bonded substance - molecule

5).  a bond characteristic of metals in which mobile valence electrons are shared among positive nuclei in the metallic crystal - metallic bond

7 0
3 years ago
What is the percent composition of nitrogen in a 2.57 g sample of Al(NO3)3?
Lisa [10]

Answer:

19.8% of Nitrogen

Explanation:

In the Al(NO₃)₃ there are:

1 atom of Al

3 atoms of N

And 9 atoms of O

The molar mass of Al(NO₃)₃ is:

1 Al * (26.98g/mol) = 26.98g/mol

3 N * (14g/mol) = 42g/mol

9 O * (16g/mol) = 144g/mol

26.98 + 42 + 144 = 212.98g/mol

We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:

2.57g * (42g/mol / 212.98g/mol) =

0.51g N

Percent composition of nitrogen is:

0.51g N / 2.57g * 100

= 19.8% of Nitrogen

6 0
2 years ago
What is the molar mass of 37.96 g of gas exerting a pressure of 3.29 on the walls of a 4.60 L container at 375 K?
Phantasy [73]

The molar mass of the gas is 77.20 gm/mole.

Explanation:

The data given is:

P = 3.29 atm,   V= 4.60 L   T= 375 K  mass of the gas = 37.96 grams

Using the ideal Gas Law will give the number of moles of the gas. The formula is

PV= nRT    (where R = Universal Gas Constant 0.08206 L.atm/ K mole

Also number of moles is not given so applying the formula

n= mass ÷ molar mass of one mole of the gas.

n = m ÷ x   ( x  molar mass) ( m mass given)

Now putting the values in Ideal Gas Law equation

PV = m ÷ x RT

3.29 × 4.60 = 37.96/x × 0.08206 × 375

15.134 = 1168.1241  ÷ x

15.134x = 1168.1241

x = 1168.1241 ÷ 15.13

x = 77.20 gm/mol

If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.

8 0
3 years ago
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