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mr Goodwill [35]
3 years ago
8

What kind of ions do metals form

Chemistry
2 answers:
maxonik [38]3 years ago
5 0

Answer:

cations

Explanation:

metals are made of postively charged ions

iVinArrow [24]3 years ago
5 0

Answer: cations

Explanation:

Metals are defined as the elements which loose electrons to attain stable electronic configuration. If an atom looses electrons, it leads to the formation of positive ions known as cations.

For Example: Sodium looses 1 electron to form Na^+ ions.

Non-metals are defined as the elements which gain electrons to attain stable electronic configuration. If an atom gains electrons, it leads to the formation of negative ions known as anions.

For Example: Fluorine gains 1 electron to form F^- ions.

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The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the
Sonbull [250]

Answer:

The rate of consumption of NH_{3} is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}

where -\frac{\Delta [NH_{3}]}{\Delta t} represents rate of consumption of NH_{3}, -\frac{\Delta [O_{2}]}{\Delta t} represents rate of consumption of O_{2}, \frac{\Delta [N_{2}]}{\Delta t} represents rate of formation of N_{2} and \frac{\Delta [H_{2}O]}{\Delta t} represents rate of formation of H_{2}O.

Here rate of formation of H_{2}O is 3.0 mol/(L.s)

From the above equation we can write-

-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}

Here \frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))

So, -\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}

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6 0
3 years ago
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The reaction is properly written as

Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃<span> (g) + 3 MgO (s) 

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Molar mass of H₂O = 18 g/mol
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Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429 

Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
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Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g

Percent Yield = Actual Yield/Theo Yield * 100
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