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Usimov [2.4K]
3 years ago
7

Based on the equations below, which metal is the least active? Pb(NO3)2(aq) + Ni (s) --> Ni(NO3)2 (aq)+ Pb(s) Pb(NO3)2(aq) +

Ag(s) --> No reaction Cu(
Chemistry
1 answer:
viva [34]3 years ago
6 0

Answer:

Ni

Explanation:

An active metal is a highly reactive metal. Active metals are found high up in the activity series.

Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.

Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.

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What is the component concentration ratio, [no2−]/[hno2], of a buffer that has a ph of 3.90? (ka of hno2 = 7.1 × 10−4)?
sasho [114]
According to Henderson–Hasselbalch Equation,

                                    pH  =  pKa + log [Nitrate] / [Nitric Acid]

As,      Ka of Nitric Acid  =  7.1 × 10⁻⁴
So,
           pKa  =  -log [ 7.1 × 10⁻⁴ ]

           pKa  =  3.148

So,                               pH  =  3.148 + log [Nitrate] / [Nitric Acid]

                                  3.90  =  3.148 + log [Nitrate] / [Nitric Acid]

                      3.90 - 3.148  = log [Nitrate] / [Nitric Acid]

                                0.752  =  log [Nitrate] / [Nitric Acid]

Taking Antilog on both sides,

                                  [Nitrate] / [Nitric Acid]  =  5.64
8 0
3 years ago
A 78 g of Aluminium bar place in water. The initical water level was 132 ml and the final water level was 144 ml. Find the Volum
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The volume is 12 mL (0.012 L)

Before the aluminum was added, the water reached 132mL. After it was added, it reached 144. That means the Aluminum takes up (144-132)= 12mL of space.

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8 0
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In which state of matter do the particles spread apart and fill all the space available to them?
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7 0
3 years ago
A total of 25.0 mL of 0.150 M potassium hydroxide (KOH) was required to neutralize 15.0 mL of sulfuric acid (H2SO4) of unknown c
8_murik_8 [283]
For neutralization of acid by a base (or vice versa), the equation should be used.
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where M's are the molarity and the Vs are the volume. Substituting the known values,
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3 0
3 years ago
Give the formula for an ionic compound formed from each pair of ions
Vera_Pavlovna [14]

Please, find the complete question in the picture attached.

Answer:

1. Na₂O

2. AlF₃

3. MgO

4. Ca₃P₂

Explanation:

1. Na⁺ and O²⁻

An ionic compound will be formed when two ions of opposite charge attract each other.

The positive ion is called cation and the negative ion is called anion.

Thus, every ionic compound has a cation and an anion electrostatically bonded.

The ions must combine in a proportion that genders a neutral compound: so you must have as many cations as negative charge has one anion and as many anions as positive charge has one cation.

This is, if the cation has charge +x, there will be x anions, and if the anion has charge -y, there will be y cations.

Simbolically:

Cation:A^{+x}\\ \\ Anion:B^{-y}\\ \\ Compound:A_yB_x

As you see, the subscripts for each element in the chemical formula are obtained by the exchage of the charges of the ions.

Then, for Na⁺ and O²⁻, the subscript for Na will be 2 and the subscript for O will be 1; and the formula of the ionic compound formed by this pair ot ions is:

  • Na₂O

2. Al³⁺ and F⁻

  • The subscript of Al will be 1 (because the F ion has charge 1-, and the subscript of F will be 3 (because the Al ion has charge +3).

Thus the ionic compound formed by this pair of ions is:

  • AlF₃

3. Mg²⁺ and S²⁻

  • The charge 2+ from Mg atom wil become the subscript 2 of S atom, and the charge 2- will become the subscript 2 of Mg atom:

That results in the formula: Mg₂S₂

Except for some special compounds, the chemical formula is simplified, dividing by the least common denominator. In this case, that means that the two 2 subscripts are simplified to 1, and the final chemical formula for the ionic compound formed by this pair of ions is:

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4. Ca²⁺ and P³⁻

  • The charge 2+ from Ca will become subscript 2 for P and the charge 3- whill become subscript 3 for Ca.

Hence, the ionic compound formed by these two ions is:

  • Ca₃P₂

7 0
3 years ago
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