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Svetllana [295]
3 years ago
14

How many valence electrons does K+ have

Chemistry
2 answers:
user100 [1]3 years ago
7 0
About one valence electron
Lera25 [3.4K]3 years ago
5 0

Answer:

About one valence electron

Explanation:

Obviously, removing that electron gives us [Ar] (same configuration as K1+), which is a noble gas and has 8 electrons. Valence electrons are generally regarded as being 'the outermost electrons' for a given atom. Therefore, with neutral potassium, there is one valence electron

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Which of the following is the energy of moving electrons?
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Electrical energy

Explanation:

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hope it helps!

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2 years ago
Which of the following is true for an atom which has 29 protons, 34 neutrons and 29 electrons?
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6 0
3 years ago
Consider the reaction
SOVA2 [1]

Answer :

(a) The average rate will be:

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

(b) The average rate will be:

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

Thus, the rate of reaction will be:

\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}

<u>Part (a) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}

and,

\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

<u>Part (b) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

and,

-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

5 0
3 years ago
OK! this is for the weebs :D
maw [93]

Answer:

Explanation:3

4 0
3 years ago
Read 2 more answers
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