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Serjik [45]
3 years ago
15

Please help!

Chemistry
1 answer:
inysia [295]3 years ago
7 0

Answer:

NH3 —> base

We will see ammonia attach itself to silver and not the other way around, silver therefore accepts electrons. So, silver is a Lewis acid and ammonia is a Lewis base, so this reaction is complex as ammonia reacts with the silver cation .

Any group that contains an empty orbital in its valence shell can be considered a Lewis acid, but the bases do not differ in the Brønsted system from those in the Lewis system. Thus, acids are divided into types, including:

Simple Cations

Theoretically, they can be considered as acids and therefore we can also expect their acidic strength to increase for the following reasons:

1. By increasing the positive charge on the cation.

2. Increasing the charge of the nucleus from one atom to another in any period in the periodic table.

3. By decreasing the radius of the cation.

4. Decrease in the number of electron shells in the cation.

Therefore, the acidity of cations of any chain of transition elements increases with increasing atomic number when their positive charge is equal. There are some examples that show the reaction of cations as Lewis acids:

A) the reaction of the silver ion with ammonia

b) aluminum ion hydration

c) Adding alcohol to the lithium ion

D) formation of a ferric cyanide ion

I hope I helped you^_^

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3 years ago
3H2 + N2 —&gt; 2NH3 .
In-s [12.5K]

Answer:

\boxed{1.26 \times 10^{20} \text{ molecules NH}_{3}}

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

M_r:      2.016                 17.03

              3H₂   +   N₂ ⟶ 2NH₃

m/g:  6.33 × 10⁻⁴  

2. Calculate the moles of H₂

\text{Moles of H}_{2} = 6.33 \times 10^{-4}\text{ g H}_{2} \times \dfrac{\text{1 mol H}_{2}}{\text{2.016 g H}_{2}} = 3.140 \times 10^{-4} \text{ mol H}_{2}

3. Calculate the moles of NH₃

The molar ratio is 2 mol NH₃/3 mol H₂.

\text{Moles of NH}_{3} = 3.140 \times 10^{-4} \text{ mol H}_{2} \times \dfrac{\text{2 mol NH}_{3}}{\text{3 mol H}_{2}} = 2.093 \times 10^{-4}\text{ mol NH}_{3}

4. Calculate the molecules of NH₃

There are 6.022 × 10²³ molecules of NH₃/1 mol NH₃.

\text{Molecules of NH}_{3} = 2.093 \times 10^{-4}\text{ mol NH}_{3} \times \dfrac{6.022 \times 10^{23}\text{ molecules NH}_{3}}{\text{1 mol NH}_{3}}\\= 1.26 \times 10^{20}\text{ molecules NH}_{3}\\\text{The reaction produces } \boxed{\mathbf{1.26 \times 10^{20}} \textbf{ molecules NH}_{3}}

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3 years ago
Find the mass of an object with a volume of 5.2cm^3 and a density of 1.3g/cm^3
Oksanka [162]

The mass of object is 6.8 g.

Mass = 5.2 cm³× (1.3 g/1 cm³) = 6.8 g

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