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3 years ago

5

All the elements in the same period have the same_______

1 answer:

gogolik [260]3 years ago

8 0

Answer:

all the elements in the same period have the same valence electrons.

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#1: Which scientist is credited with developing the orbital model of the atom?

CaHeK987 [17]

<span>Niels Henrik David Bohr is the scientist credited with developing the orbital model of the atom. He was Danish physicist who made foundational contributions to understanding atomic structure and quantum theory. He received the Nobel Prize in Physics in 1922</span>

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3 years ago

What is the pH of a solution with a concentration of 4.7 × 10-4 molar H3O+?

wlad13 [49]

The pH of a liquid substance is calculated through the equation,
pH = -log[H3O+]
Substituting the given concentration of the hydronium ion to the equation above,
pH = -log[4.7x10^-4 M]
The value of pH is equal to 3.33. Thus, the pH of the solution is approximately 3.33.

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3 years ago

Which statement(s) is/are TRUE about covalent bonds?

tresset_1 [31]

Answer:

1. Covalent bonds can form between two nonmetal atoms.

2. Covalent bonds can form between atoms of the same element.

3. Covalent bonds can form between atoms of different elements.

Explanation:

I hope this helps u! :D

Explanation:

3 0

3 years ago

The carton fe3+ is formed when

vekshin1

Answer:

The cation Fe3+ is formed when a. an atom of iron loses two electrons.

Explanation:

7 0

3 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago