Answer:
The elements in the periodic table are arranged in order of increasing atomic number. All of these elements display several other trends and we can use the periodic law and table formation to predict their chemical, physical, and atomic properties. Understanding these trends is done by analyzing the elements electron configuration; all elements prefer an octet formation and will gain or lose electrons to form that stable configuration.
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How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?
Answer:
[tex]m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2[/tex]
Explanation:
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In this case, since the chemical reaction between copper and nitric acid is:
[tex]2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2[/tex]
By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:
[tex]m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2[/tex]
However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:
[tex]4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2[/tex]
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Amount of copper (II) nitrate would be produced is 2.25 gram
Molecular mass:Given that;
Amount of copper metal = 0.80 g
Find:
Amount of copper (II) nitrate would be produced
Computation:
Cu + 4HNO₃ → Cu(NO₃) + 2NO₂ + 2H₂O
Number of moles = Given mass / Molecular mass
Number of moles = 0.80 / 63.5
Number of moles = 0.012 mol
So,
n = Given mass / Molecular mass
0.012 = Given mass / 63.5 + (63.5 + 48)
Given mass = 2.25 gram
Amount of copper (II) nitrate would be produced = 2.25 gram
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Choose the formula for ammonium.
ΝΗ,
b NH3
C
(NH)3N
Answer:
NH4+ will be ammonium, NH3 is ammonia which is the conjugate base of ammonium.
Explanation: