What type of elements are in s-block

1 answer:

4 0

There are 2 different types of elements is s-block.
1. Alkali earths metals.

2. Alkaline earths metals.

2 Exceptions are Hydrogen and helium gas.

Examples: Sodium, potassium, francium, beryllium, magnesium... etc

You might be interested in

Which of the following weak interactions would require the least amount of energy to disrupt? Group of answer choices

SpyIntel [72]

Answer: Option (c) is the correct answer.

Explanation:

A hydrogen bond is defined as a weak bond that is formed between an electropositive atom (generally hydrogen atom) and an electronegative atom like oxygen, nitrogen and fluorine.

An ionic bond is defined as a bond formed between a metal and a non-metal and in this bond transfer of electron takes place from metal to non-metal. And, due to the presence of opposite charges on the combining atoms there exists a strong force of attraction.

Vander waal forces are defined as the weak electric forces which tend to attract neutral molecules towards each other in gases, liquefied and solidified gases.

Vander waal forces are very weak forces.

Thus, we can conclude that Van der walas interactions are weak interactions would require the least amount of energy to disrupt.

8 0

2 years ago

Read 2 more answers

Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

Illusion [34]

Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$