The moles of I₂ will form from the decomposition of 3.58g of NI₃ is 0.0136 moles.
<h3>How we calculate moles?</h3>
Moles of any substance will be calculated as:
n = W/M, where
W = required mass
M = molar mass
Given chemical reaction is:
2NI₃ → N₂ + 3I₂
Moles of 3.58g of NI₃ will be calculated as:
n = 3.58g / 394. 71 g/mol = 0.009 moles
From the stoichiometry of the solution, it is clear that:
2 moles of NI₃ = produce 3 moles of I₂
0.009 moles of NI₃ = produce 3/2×0.009=0.0136 moles of I₂
Hence, option (3) is correct i.e. 0.0136 moles.
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Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).
What is standard entropy?
The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."
Calculation:
Balancing the given reaction following-
1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)
ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]
Here S° = standard entropy of the system
Insert into the aforementioned equation all the typical entropy values found in the literature:
ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]
⇒ΔS° = - 99.4 J/K
Therefore, the standard entropy, ΔS° is -99.4 J/K.
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Imagine we have <span>mass of solvent 1kg (1000g)
According to that: </span>

= 4.8 mole * 98 g/mole = 470g


m(H2SO4) which is =<span>470g
</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>
Answer:
english pls so i can answer
Ion-dipole forces
H2O has hydrogen bonding, which is a form of dipole-dipole forces, and NO3- is an ion, so the intermolecular attraction is ion-dipole.