Answer:
18.06 × 10²³ molecules
Explanation:
Add the two amounts of molecules together.
6.02 × 10²³ + 12.04 × 10²³ = 18.06 × 10²³
You will have 18.06 × 10²³ molecules in the vessel when the reaction is complete. This is because of the Law of Conservation of Mass. Mass is neither created nor destroyed in chemical reactions. You will have the exact number of molecules in the reaction vessel as you did in the beginning. The types of molecules may change, but the number will stay the same.
Answer:
Explanation:
Potassium nitrate is a soluble salt which readily dissolves in a polar solvent, such as water. When solid potassium nitrate is dissolved in water, it dissociates into potassium cations and nitrate anions.
Due to the resultant ionic charges, the polar water molecules attract the resultant ions and potassium nitrate ions become hydrated, that is, surrounded by water molecules.
Nitrate, our anion, attracts the partially positive ends of water molecules by attracting them via hydrogen atom.
Potassium, the cation, attracts the partially negative end of water molecules by attracting via oxygen atom.
Calcium has the greatest mass of the first twenty elements with a mass of approximately 40.08
I'll see what I can do here...
1) Nonmetal
2) Calcium (Ca), chemical element, one of the alkaline-earth metals of Group 2 (IIa) of the periodic table.
3) Hafnium
4) 204.3833 u
5) Not sure what you're asking, but oble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)
6) The metalloids; boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At)
7) The Actinide series contains elements with atomic numbers 89 to 103 and is the third group in the periodic table.
8) 33
9) 88
10) 30
Hope this helps!
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is
