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3 years ago

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

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3 years ago

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horrorfan [7]

The reaction is properly written as

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

Molar mass of Mg</span>₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol

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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Theo Yield = 4.57 g

Percent Yield = Actual Yield/Theo Yield * 100
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