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Sav [38]
3 years ago
14

How many moles of NaCl will be produced from 83.0g of Na, assuming Cl2 is available in excess?

Chemistry
2 answers:
SashulF [63]3 years ago
5 0

Answer:

2.57

Explanation:

Sindrei [870]3 years ago
3 0

Answer: The number of moles of NaCl produced will be, 3.61 moles.

Explanation : Given,

Mass of Na = 83.0 g

Molar mass of Na = 23 g/mol

First we have to calculate the moles of Na

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{83.0g}{23g/mol}=3.61mol

Now we have to calculate the moles of NaCl

The balanced chemical equation is:

2Na+Cl_2\rightarrow 2NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 3.61 mole of Na react to give 3.61 mole of NaCl

Therefore, the number of moles of NaCl produced will be, 3.61 moles.

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4 0
2 years ago
The bonding of two atoms will most likely occur if:
sergejj [24]
The bonding of two atoms will most likely occur if A. a more stable state can result from the union.
Otherwise, there would be no point in making two atoms bond.
7 0
3 years ago
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if you were to compare the mass of the products and reactants in a reaction, you would find that the mass of the products is alw
Salsk061 [2.6K]
If you were to compare the mass of the products and reactants in a reaction, you would find that the mass of the products is <span>equal to the mass of the reactants.</span>
3 0
3 years ago
Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -
kogti [31]
The given complex ion is as follow,

                                              [Ru (CN) (CO)₄]⁻

Where;
            [ ]  =  Coordination Sphere

            Ru  =  Central Metal Atom  =  <span>Ruthenium

            CN  =  Cyanide Ligand

            CO  =  Carbonyl Ligand

The charge on Ru is calculated as follow,

                               Ru + (CN) + (CO)</span>₄  =  -1
Where;
            -1  =  overall charge on sphere

             0  =  Charge on neutral CO

            -1  =  Charge on CN

So, Putting values,


                               Ru + (-1) + (0)₄  =  -1

                               Ru - 1 + 0  =  -1

                               Ru - 1  =  -1

                               Ru  =  -1 + 1

                               Ru  =  0
Result:
          <span>Oxidation state of the metal species in each complex [Ru(CN)(CO)</span>₄]⁻ is zero.
3 0
3 years ago
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