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crimeas [40]
2 years ago
5

It is easier to determine the electron configurations for the p-block elements in periods 1, 2, and 3 than to determine the elec

tron configurations for the rest of the p-block elements in the periodic table because
they have a larger number of core electrons.
their electrons are assigned to s and p orbitals only.
their electrons are placed in a higher number of orbitals.
they have more valence electrons available for bonding.
Chemistry
1 answer:
babymother [125]2 years ago
5 0

Answer:

Their electrons are placed in a higher number of orbitals

Explanation:

  • Suppose a element be Ga .

The atomic no is 31

The configuration is given by

\\ \sf\longmapsto 1s^22s^22p^63ss^23p^63d^{10}4s^24p^1

Or

\\ \sf\longmapsto [Ar]3d^{10}4s^24p^1

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Which element would you expect to be more metallic?<br> (a) Ca or Rb (b) Mg or Ra (c) Br or I
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Explanation:

When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.

As calcium (Ca) is a group 2A element and rubidium (Rb) is a group 1A element. Hence, Rb being an alkali metal is more metallic in nature than calcium (alkaline earth metal).

Both magnesium (Mg) and radium (Ra) are group 2A elements. And, when we move down a group then as the size of element increases so, it becomes easy of the metal atom to lose an electron.

As a result, there occurs an increase in metallic character of the element. Hence, Radium (Ra) is more metallic in nature than magnesium (Mg).

Also, both bromine and iodine are group 17 elements. Since, both of them are non-metals and non-metallic character increases on moving down the group.

Therefore, bromine (Br) is more metallic than iodine.

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3 years ago
A gas of 19 mL at a pressure of 740 mmHg can be expected to change its pressure when its volume changes to 30. mL. Express its n
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To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

 

P1V1 =P2V2

P2 = P1V1/V2

P2 = 740mmhg x 19 mL / 30 mL

<span>P2 = 468.67 mmHg = 0.62 atm</span>

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If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?
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<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol

  • <u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol

Volume of solution = 50 + 60 = 110 mL = 0.11 L    (Conversion factor:  1 L = 1000 mL)

  • To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of TRIS acid = 8.3

[\text{TRIS acid}]=\frac{0.005}{0.11}

[\text{TRIS base}]=\frac{0.012}{0.11}

pH = ?

Putting values in above equation, we get:

pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7

Hence, the pH of resulting solution is 8.7

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3 years ago
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