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konstantin123 [22]
3 years ago
14

How many valence electrons are there in sulfur, and iodine.

Chemistry
1 answer:
Vinil7 [7]3 years ago
7 0

Today I will explain what a valence electron is and how many valence electrons each of the atoms your asking about have. :)

→ A valence electron is an electron in the outer shell of an atom that is able to participate in bonding. While iodine has 53 electrons, all but seven are in the inner shells of the iodine atom.

→ Sulfur has sixteen electrons in total, two in the inner shell, eight in the middle, and six in the outer. This means that sulfur has six valence electrons.

↑   ↑   ↑   Hope this helps! :D

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The Weak Ionization constant (Ka) for HCOOH is equal to:<br><br><br> Please help me lol
iris [78.8K]

Answer:

Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}

Explanation:

Hello,

In this case, the weak ionization reaction for formic acid is:

HCOOH(aq)\rightleftharpoons H^+(aq)+HCOO^-(aq)

In such a way, we simply recall the law of mass action in order to represent the weak ionization constant, Ka, for such process, by taking into account that the concentration of products is divided over the concentration of reactants as shown below:

Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}

Best regards.

8 0
3 years ago
Read 2 more answers
Pls help need this to pass
SOVA2 [1]

Bulk property  are those properties that does not depend on the amount of substance.

<h3>What is a bulk property?</h3>

The term bulk property can also be referred to as the intensive properties of a substance. These are those properties that does not depend on the amount of substance. This means that they are characteristic of the material.

The percent by mass refers to  the percentage of an atom that is present in a compound. If we must be able to calculate the percentage by mass then we must know the mass of the sample and the masses of the elements found in the sample.

Learn more about bulk properties:brainly.com/question/25960022

#SPJ1

8 0
2 years ago
Read 2 more answers
Which of the following will increase the rate of disolving?
Fantom [35]

Answer:

I have no idea bro

Explanation:

I feel you

3 0
3 years ago
(i) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.
Leni [432]

Answer:

(c)(i) The order of the reaction based on the graph provided is first order.

(ii) 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

(ii) Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there's only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

6 0
3 years ago
What changes in space result from the death of massive stars?
lana [24]

Whenever the fuel is being used up, a star explodes and the energy leakage from a star's core ceases.

Explanation:

The dying star expands in the "Red Giant," before even the inevitable collapse starts, due to nuclear reactions just outside of the core.

It becomes a white dwarf star when the star has almost the same density as the Sun. If it's much larger, a supernova explosion could take place and leave a neutron star away. However, if it is very large–at least three times the Sun's mass–the crumbling core of the star, nothing will ever stop it from crumbling. The star is imploding into a black hole, an endless gravitational loop in space.

6 0
3 years ago
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