Answer:
The balanced equation is 2K(s) + Cl2(g)→2KCl(s)
Answer:
b. The number of electrons
Explanation:
A "neutral atom" has a <u>neutral charge</u>. This means that <em>its charge is equal to </em><em>zero. </em>In order for the charges to cancel out each other, the atom's <em>positive charge should be equal to the negative charge. </em>These being said, the number of electrons<em> (negatively-charged)</em> is then equal to the number of protons <em>(positively-charged). </em>Those atoms which are not neutral are called <em>"ions."</em> This means that they either have more or less electrons than the protons.
Answer:
The charged carbon atom of a carbocation has a complete octet of valence shell electrons
Explanation:
A charged carbon atom of a carbocation has a valence shell that is not filled, <u>that's why it acts as an electrophile (or a Lewis base)</u>. This unfilled valence shell is also the reason of the nucleophilic attack that takes place during the second step of a SN1 reaction.
Answer:
The correct answer is no.
Explanation:
Tellurium is a chemical element denoted by Te and having atomic number 52. It is mildly toxic, brittle, silver-white, and rare metalloid. The element is chemically related to sulfur and selenium, all three of which are chalcogens.
Oxygen is a chemical element, that is, a substance, which comprises only one kind of atom. Its official chemical symbol is O and exhibits an atomic number 8, this signifies that an atom of oxygen possesses eight protons in its nucleus. In the given question, it is not likely that tellurium would replace for oxygen, as the two elements are highly unlike.
Explanation:
(a) As the given chemical reaction equation is as follows.
So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.
Hence, equation for rate law of reaction will be as follows.
Rate = ![K \times [OCl^{-}] \times [l^{-}]](https://tex.z-dn.net/?f=K%20%5Ctimes%20%5BOCl%5E%7B-%7D%5D%20%5Ctimes%20%5Bl%5E%7B-%7D%5D)
(b) Since, the rate equation is as follows.
Rate = ![K [OCl^{-}][l^{-}]](https://tex.z-dn.net/?f=K%20%5BOCl%5E%7B-%7D%5D%5Bl%5E%7B-%7D%5D)
Let us assume that (![[OCl^{-}] = [l^{-}]](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%20%5Bl%5E%7B-%7D%5D)
)
Putting the given values into the above equation as follows.
K = 
= 
Hence, the value of rate constant for the given reaction is
.
(c) Now, we will calculate the rate as follows.
Rate =
= 
= 
Therefore, rate when ![[OCl^{-}] = 1.8 \times 10^{3}](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%201.8%20%5Ctimes%2010%5E%7B3%7D)
M and ![[I^{-}]= 6.0 \times 10^{4}](https://tex.z-dn.net/?f=%5BI%5E%7B-%7D%5D%3D%206.0%20%5Ctimes%2010%5E%7B4%7D)
M is .
