Given the reaction _K(s) +_ Cl2(g) → _KCl(s) what is the amount of K, in grams, needed to completely react with 2 moles of Cl2(g
1 answer:
Answer:
156.4g K
Explanation:
I'm not sure if it is correct but I think it should be this
What do we know so far?: 2K + 1Cl2 -> 2KCl, 2 mol of Cl2
What are we looking for?: #g of K
What is the ratio of K to Cl2?: 2:1
Set up equation: 2molCl2 x
Cancel unwanted units: 2 x
Answer we got: 2 x 2mol K = 4mol K
Converting moles to grams: 4 x 39.1 (molar mass of K) = 156.4g K
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<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61
<u>Explanation:</u>
We are given:
Initial moles of iodine gas = 0.100 moles
Initial moles of hydrogen gas = 0.100 moles
Volume of container = 1.00 L
Molarity of the solution is calculated by the equation:
Equilibrium concentration of iodine gas = 0.0210 M
The chemical equation for the reaction of iodine gas and hydrogen gas follows:
<u>Initial:</u> 0.1 0.1
<u>At eqllm:</u> 0.1-x 0.1-x 2x
Evaluating the value of 'x'
The expression of for above equation follows:
Putting values in above expression, we get:
Hence, the value of equilibrium constant for the given reaction is 56.61
Answer:
See explanation below (Brainlist please)
Explanation:
First of all, we need to understand what is an acidic hydrogen.
An acidic hydrogen, is the atom of hydrogen which is more propense to undergo an acid base reaction, and form a stable ion or molecule in the process.
In other words, is the hydrogen that is more vulnerable to get substracted in an acid base reaction to form another compound.
Knowing this information, gives us an idea of how a molecule can be formed and which kind of compound is formed.
Now, in this question, we have 5 molecules. Each of them is either a ketone or aldehyde, so this mean that we have the carbonile group (C = O), which means that is easier to identify the acidic hydrogen. This is because the Carbonile group is an attractor group, so, it will attract the charges by inductive effect (in some cases by resonance), and the molecule is more stable.
This can be shown by drawing the enolate ion that is formed once the molecule undergo the acid base reaction. As it's an enolate form that we are looking for, then it means that the ketone or aldehyde is undergoing an electrofilic attack with a base. This base will substract the most acidic hydrogen to form a better and stable enolate. The acidic hydrogen and the enolate form can be seen in the attached picture.
a) In the case of acetaldehyde, the most acidic will be the hydrogen of carbon 2, because the hydrogen from the carbonile, once it's substracted, the charge of the carbon cannot be stabilized by resonance. Carbon 2 hydrogens, can do this job easily.
b) Propanal happens something similar to acetaldehyde, the terminal hydrogen cannot be substracted, and carbon 3, once the hydrogen is gone, the negative charge cannot be stabilized by resonance, so hydrogens of carbon 2 can do this.
c) in the case of acetone, is easier to look because we only have the C = O between two methyl group, so you can use either carbon 1 or 3 to do the job.
d) 4 heptanone the most acidic hydrogen would be carbon 3 or 5, because they are closer to the C=O and the ion can be stabilized by resonance.
e) Finally in ciclopentanone, the most acidic hydrogen would be carbon 2 or 5.
See picture for a better understanding.
Hope it helps.
