Answer:
<u>One lone-Pair is present in Ammonia</u>
<u></u>
Explanation:
The number of valence electron in N = 5
The number of Valence electron in H = 1
The formula of ammonia = NH3
Total valence electron in ammonia molecule = 5 +3(1) = 5+3 = 8
The lewis structure suggest that :
Nitrogen completes its octet by sharing the electron pair with 3 hydrogen atoms.
3 electron of Nitrogen are involved in sharing with Hydrogen
So,<u><em> remaining two electron are left non-bonded</em></u> . Hence they exist as lone- pair
So, there is only 1 lone pair in the ammonia molecule .
The shape of NH3 is bent according to VSEPR theory . This is so because the presence of 1 lone pair causes more repulsion and occupy more space.
Thus the lone pair is changing the shape of the ammonia molecule . It also increase the dipole moment of the molecule , which gives polarity to it.
Answer:
569K
Explanation:
Q = 3.5kJ = 3500J
mass = 28.2g
∅1 = 20°C = 20 + 273 = 293K
∅2 = x
c = 0.449
Q = mc∆∅
3500 = 28.2×0.449×∆∅
3500 = 12.6618×∆∅
∆∅ = 3500/12.6618
∆∅ = 276.4220
∅2 - ∅1 = 276.4220
∅2 = 276.4220 + ∅1
∅2 = 276.4220 + 293
∅2 = 569.4220K
∅2 = 569K
i think the greater the electric charge the atom decreases in size
The correct answer is species
As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its 
can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.
Let the three equations with given be denoted as (1), (2), (3), and the last equation (4). Let 
, 
, and 
be letters such that 
. This relationship shall hold for all chemicals involved.
There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 
shall resemble the number of 
left on the product side when the second equation is directly added to the third. Similarly
Thus
and
Verify this conclusion against a fourth species involved- 
for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.
Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.
