All categories

3 years ago

All p block elements form ions. True or False

Chemistry

1 answer:

FinnZ [79.3K]3 years ago

4 0

Answer:

No this statement is false.

Explanation:

when an atom lose or gain the electron ions are formed. If the atom lose an electron the positive ions are formed called cations while anion is formed by the gaining of electron by an atom The elements having less electrons in valance shell usually lose their electrons while the elements like halogens having 7 valance electrons gain an electron to complete the octet. while p block elements noble gases are inert, their outer most valance shell is complete so they do not form ions.

P block elements are non-metals, metals and metalloids. These are thirty five elements. The P-block elements are present on right side of periodic table. There valance electrons are present in P orbital. The p-block metals are shiny and good conductor of heat and electricity. These metal lose the electron which is accept by non metals and form ionic bond. They have high melting points.

Metalloids includes boron, silicon, germanium, arsenic, antimony and tellurium. Metalloids contain both the properties of metals and non metals, Some metalloids are toxic like arsenic.

Most of p-block elements are non metals. They are bad conductor of heat and electricity and have low boiling points. The non metals mostly accept the electron from the metals and usually from ionic bond like in case of chlorine. It form the ionic compound with sodium.

The sodium chloride which is an ionic compound, formed by the complete transfer of electron from sodium to chlorine atom and form ionic bond. In this ionic compound sodium carry positive charge and chlorine carry negative charge there is attraction between these oppositely charged atoms.

You might be interested in

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

_________________________________
_________________________________

Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

3 years ago

Calculate the pH of mixing 24 mL of 1M acetic acid with 76 mL of 1M sodium acetate. For the purpose of this calculation, assume

FromTheMoon [43]

Answer:

pH = 5.24

Explanation:

Mixture of acetic acid with acetate ion is a buffer (Mixture of a weak acid with its conjugate base). The pH of a buffer can be determined using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

Where pKa is -log Ka = 4.74; [A⁻] is the concentration of conjugate base (Acetate ion) and [HA] is molar concentration of the weak acid.

Concentration of the acetic acid in the 100mL≡0.1L (76mL + 24mL) solution is:

[HA] = 0.024L ₓ (1mol / L) / 0.1L = 0.24M

[A⁻] = 0.076L ₓ (1mol / L) / 0.1L = 0.76M

Replacing in H-H equation:

pH = 4.74 + log₁₀ [0.76M] / [0.24M]

7 0

3 years ago

Complete the given equation and mention whether heat is absorbed or evolved during the chemical reaction.

DaniilM [7]

Answer:

NaCI + H2O

this is answer

5 0

2 years ago

The concentration of a solution can be changed by _______ solute.

Snezhnost [94]

Add more solute..............

7 0

3 years ago

Which of the following is true about the efficiency of energy transfer in an ecosystem?

Firlakuza [10]

Answer:

C. The less energy the organism requires, the more efficient the energy transfer.

8 0

3 years ago