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Greeley [361]
2 years ago
13

Identify the Lewis acid and Lewis base from among the reactants in each of the following equations.

Chemistry
1 answer:
Alex2 years ago
3 0

Answer:

Part A

Ag+ is the Lewis acid and NH3 is the Lewis base.

Part B

AlBr3 is the Lewis acid and NH3 is the Lewis base.

Part C

AlCl3 is the Lewis acid and Cl− is the Lewis base.

Explanation:

A Lewis acid is any specie that accepts a lone pair of electrons. Ag^+, AlBr3 and AlCl3 all accepted lone pairs of electrons according to the three chemical reaction equations shown. Hence, they are Lewis acids.

A Lewis base donates a lone pair of electrons. They include neutral molecules having lone pair of electrons such as NH3 or negative ions such as Cl- .

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Find the mass of 5.82 mol of MgCl2. Round to the nearest tenth.
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In a room full of air, the air is mainly composed of Nitrogen and Oxygen molecules (both at room temperature). Find (to two sign
zloy xaker [14]

Explanation:

Expression for the v_{rms} speed is as follows.

            v_{rms} = \sqrt{\frac{3kT}{M}}

where,   v_{rms} = root mean square speed

                     k = Boltzmann constant

                    T = temperature

                    M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of v_{rms} as follows.

               v_{rms} = \sqrt{\frac{3kT}{M}}

                            = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}

                            = 498.5 m/s

Hence, v_{rms} for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its v_{rms} speed as follows.

                v_{rms} = \sqrt{\frac{3kT}{M}}

                              = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}

                              = 524.5 m/s

Therefore, v_{rms} speed for nitrogen is 524.5 m/s.

3 0
3 years ago
What mass of HgO is required to produce 0.692 mol of O2?<br><br>2HgO(s) -&gt; 2Hg(l) + O2(g)
Vika [28.1K]

The answer for the following problem is mentioned below.

  • <u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>

Explanation:

Given:

no of moles of the oxygen gas = 0.692

Also given:

2 HgO  → 2 Hg + O_{2}

where,

HgO represents mercuric oxide

Hg represents mercury

O_{2} represents oxygen

To calculate:

Molar mass of HgO:

Molar mass of HgO = 216 grams

molar mass of mercury (Hg) = 200 grams

molar mass of oxygen (O) =16 grams

HgO = 200 +16 = 216 grams

We know;

       2×216 grams of HgO   →  1 mole of oxygen molecule

             ?                              →  0.692 moles of oxygen molecule

       

          = \frac{2*216*0.692}{1}

      = 298.944 grams of HgO

<u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>

<u />

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