A) eventually they will be in thermal equilibrium
Answer:
5.2g copper (Cu) => 0.082 moles copper (2 sig.figs.)
Explanation:
mole conversions:
grams to moles => divide by formula wt.
moles to grams => multiply by formula wt.
gas volumes to moles => divide volume by 22.4Liters/mole (STP conditions only)
This problem:
mass to moles => divide by formula wt.
mass = 5.2g = 5.2g/63.5g/mole = 0.082 moles copper (2 sig.figs.)
Answer:
The correct answer is "transferred; unequally shared; equally shared".
Explanation:
Ionic bonding occurs when a positively charged atom (cation) interacts with a negatively charged atom (anion). In ionic bonding, the cation transfers its electron to the anion. In polar covalent bonding, electrons are unequally shared. This means that the electrons spend more time in an atom than the other, which gives partial positive and negative charges to the atoms. On the other hand in nonpolar covalent bonding, the electrons are equally shared and no charges are created.
Answer:
Explanation:
For a first order reaction the rate law is:
![v=\frac{-d[A]}{[A]}=k[A]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B-d%5BA%5D%7D%7B%5BA%5D%7D%3Dk%5BA%5D)
Integranting both sides of the equation we get:
![\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%7D%20%5C%2C%20dx%20%3D-k%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt)
where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.
From that integral we get the integrated rate law:
![ln\frac{[A]}{[A]_{0} } =-kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_%7B0%7D%20%7D%20%3D-kt)
![[A]=[A]_{0}e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_%7B0%7De%5E%7B-kt%7D)
![ln[A]=ln[A]_{0} -kt](https://tex.z-dn.net/?f=ln%5BA%5D%3Dln%5BA%5D_%7B0%7D%20-kt)
![k=\frac{ln[A]_{0}-ln[A]}{t}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln%5BA%5D_%7B0%7D-ln%5BA%5D%7D%7Bt%7D)
therefore k is
