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Archy [21]
3 years ago
7

18 g of copper is mixed with silver nitrate in water. How much copper ll nitrate will formed?

Chemistry
1 answer:
Lorico [155]3 years ago
7 0

Answer:

Mass = 112.54 g

Explanation:

Given data:

Mass of copper = 18 g

How much copper(II) nitrate formed = ?

Solution:

Cu + 2AgNO₃  →  Cu(NO₃)₂ + 2Ag

Number of moles of copper:

Number of moles = mass/ molar mass

Number of moles = 18 g/ 29 g/mol

Number of moles = 0.6 mol

Now we will compare the moles of Cu with Cu(NO₃)₂ .

             Cu        :        Cu(NO₃)₂

              1           :             1

           0.6          :           0.6

Mass of Cu(NO₃)₂ :

Mass = number of moles × molar mass

Mass = 0.6 mol × 187.56 g/mol

Mass = 112.54 g

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Leviafan [203]

Answer:

 4.90  moles of  Mg(ClO_4)_2   will produce  (9.8) moles of  Cl^{-} ,

                                                    (4.90) moles of Mg^{2+} and

                                                    (39.2) moles of  O^{2-}

Explanation:

From the question we are told that

  The number of moles of  is  n =  4.90  \ mols

The formation reaction of Mg(ClO_4)_2  is

             Mg^{2+} + 2 Cl^{-} + 8O^{2+} \to Mg(ClO_4)_2  

From the reaction we see that

   1 mole of  Mg(ClO_4)_2  is formed by 2 moles of Cl^{-} 1 mole of  Mg^{2+} and 4  O^{2-}

 This implies that

   4.90  moles of  Mg(ClO_4)_2   will produce  (2 * 4.90) moles of  Cl^{-} ,

                                                    (1 * 4.90) moles of Mg^{2+} and

                                                    (8 * 4.90) moles of  O^{2-}

So

  4.90  moles of  Mg(ClO_4)_2   will produce  (9.8) moles of  Cl^{-} ,

                                                    (4.90) moles of Mg^{2+} and

                                                    (39.2) moles of  O^{2-}

                           

4 0
3 years ago
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3 years ago
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stiks02 [169]

Answer:

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Explanation:

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4 0
3 years ago
while doing a lab, a student found the density of a piece of pure aluminum to be 2.85 g/cm^3. the accepted value for the density
vagabundo [1.1K]

Answer:

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Explanation:

Given data:

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Formula:

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Now we will put the values in formula:

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Percent error = [0.15/2.70 g/cm³ ] × 100

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Percent error = 5.6%

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