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kicyunya [14]
1 year ago
6

Sodium is an alkali metal that reacts violently with water. Calculate the total number of atoms in a 58.2 grams sample of sodium

and express in correct scientific notation.
Chemistry
1 answer:
Evgesh-ka [11]1 year ago
4 0

In 58.2 grams of Na there is 1.51 x 10^24 atoms.

<h3>Mole calculation</h3>

To calculate the amount of atoms present in a mass sample, it is necessary to have knowledge of the molar mass of the element and Avogadro's constant, used in such a way that:

MM_Na= 23g/mol

C = 6\times 10^{23}

                                       MM = \frac{m}{mol}

                                    23 = \frac{58.2}{x} = > 2.53 atoms

Now, with the value of the number of moles, just calculate the number of atoms present, so that:

                                       atoms = mol \times C

                                     atoms = 2.53 \times 6 \times 10^{23}

                                        atoms = 1.51 \times 10^{24}

So, in 58.2 grams of Na there is 1.51 x 10^24 atoms.

Learn more about mole calculation in: brainly.com/question/2845237

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The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

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Explanation:

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Volume of solution = 1.00  L

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The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

Now put all the given values in this expression, we get :

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

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