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Dennis_Churaev [7]
3 years ago
15

How many atoms are in 16.67g of Calcium Nitrate?

Chemistry
2 answers:
Jlenok [28]3 years ago
5 0
<span>5.506x10^23 atoms First, determine the molar mass of Calcium Nitrate [ Ca(NO3)2 ]. Look up the atomic weights of the involved elements. Atomic weight calcium = 40.078 Atomic weight nitrogen = 14.0067 Atomic weight oxygen = 15.999 Molar mass Ca(NO3)2 = 40.078 + 2 * 14.0067 + 6 * 15.999 = 164.0854 g/mol Determine the number of moles of Ca(NO3)2. 16.67 g / 164.0854 g/mol = 0.101593439 mol Now multiply by the number of atoms in Ca(NO3)2 which is 9. So 9 * 0.101593439 = 0.914340947 Finally, multiply by avogadro's number. Giving: 0.914340947 * 6.0221409x10^23 = 5.50629001x10^23 Round the result to 4 significant figures, giving 5.506x10^23 atoms.</span>
Marta_Voda [28]3 years ago
4 0
The number of atoms in calcium nitrate can easily be found by multiplying a given mole of Calcium Nitrate by Avogadro's number since each mole of a substance contains 6.03 × 10²³.

Now, mol of Ca(NO₃)₂ = mass of Ca(No₃)₂ ÷ molar mass of Ca(NO₃)₂
          mol of Ca = 16.67 g ÷ [(40 × 1) + (14 × 2) + (16 × 3 × 2) g/mol
          mol of Ca = 0.1016 mol

⇒ atoms of Ca(NO₃)₂ = 0.0812 mol × (6.02 × 10²³ atoms/mole)
                                    = 6.12 × 10²² atoms  
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3 years ago
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The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
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Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

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