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AlladinOne [14]
2 years ago
15

An aqueous solution of ammonium sulfate is allowed to react with an aqueous solution of calcium nitrate.

Chemistry
1 answer:
Salsk061 [2.6K]2 years ago
5 0

<u>Answer:</u> The net ionic equation contains Ca^{2+}(aq.) ions

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

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.

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

(NH_4)_2SO_4(aq.)+Ca(NO_3)_2(aq.)\rightarrow CaSO_4(s)2NH_4NO_3(aq.)

Ionic form of the above equation follows:

2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ca^{2+}(aq.)+2NO_3^{-}(aq.)\rightarrow CaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonia 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:

Ca^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)

Hence, the net ionic equation contains Ca^{2+}(aq.) ions

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What is the formula for manganese (ii) fluoride decahydrate? (you may use a * to represent the dot in the formula.)?
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These types of molecules are called hydrates. They have a certain number of moles attached to the salt. Their characteristic is being hygroscopic. That means that when they are exposed to air, they readily solvate. 

The formula for Manganese Fluoride Decahydrate will involve the formula Mn, F and H₂O. In ionic form, Manganese is Mn⁺² while fluoride is in F⁻. When they are brought together, their superscripts are 'cross-multiplied' and becomes their respective subscripts. The compound becomes MnF₂. Then, we add the decahydrate which means 10 moles of H₂O. Hence, the formula for Manganese Fluoride Decahydrate is MnF₂*10H₂O.
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A sodium ion has a radius of 1.16 x 10^-10 m and a nearby fluoride ion has a radius of 1.9 x 10^-10 m. Determine the distance be
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The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.

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d(NaF) = r(Na⁺) + r(F⁻).

d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.

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The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.

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3 years ago
How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl
VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3}  }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}

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