Sodium is a metal, Chloride is a non-metal.
Right off the bat, you know that in order for both of these atoms to achieve a full valence shell that the metal has to lose electrons, and the non-metal has to gain them.
Therefore, you have the transfer of electrons in this bond in order to form ions.
Na+ and Cl-. This transfer of electrons in a bond is called an {{ Ionic Bond}}
<h2>The required "option is b) hydrogen bonds must be broken to raise its temperature.</h2>
Explanation:
- Water has high specific heat due to hydrogen bonds present in it.
- The Ionisation of water does not affect the specific heat of the water.
- On decreasing the temperature, there is the formation of bonds hence option (d) is wrong.
- On increasing the temperature, there is the breaking of bonds hence option (b) is correct.
Answer:

Explanation:
The number of valence electrons tells us the group number of the neutral atom.
The atom has 4 valence electrons.
The atom is in group 4.
Answer:
7.98 × 10^3grams.
Explanation:
To find the mass of fluorine in the number of atoms provided, we first divide the number of atoms by Avagadros number (6.02 × 10^23atoms) to get the number of moles in the fluorine atom. That is;
number of moles (n) = number of atoms (nA) ÷ 6.02 × 10^23 atoms
n = 2.542 × 10^26 ÷ 6.02 × 10^23
n = 0.42 × 10^ (26-23)
n = 0.42 × 10^3
n = 4.2 × 10^2moles
Using mole = mass ÷ molar mass
Molar/atomic mass of fluorine (F) = 19g/mol
mass = molar mass × mole
Mass (g) = 19 × 4.2 × 10^2
Mass = 79.8 × 10^2
Mass = 7.98 × 10^3grams.
a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)

R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide