Answer:
D.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.
Explanation:
For the reaction:
3H₂(g) + N₂(g) → 2NH₃(g)
The enthalpy change is ΔH = -92kJ
This enthalpy change is defined as the enthalpy of products - the enthalpy of reactants. As the enthalpy is <0, The enthalpy of products is <em>lower </em>than the enthalpy of reactants.
Also, it is possible to obtain the enthalpy change from the bond energies of products - bond energies of reactants, thus, The total bond energies of products are <em>lower</em> than the total bond energies of reactants.
The rate of the reaction couldn't be determined using ΔH.
As the bond energy of ammonia is lower than bonds of nitrogen and hydrogen, <em>D. Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.</em>
I hope it helps!
Answer:
-3.82ºC is the freezing point of solution
Explanation:
We work with the Freezing point depression to solve the problem
ΔT = m . Kf . i
ΔT = Freezing point of pure solvent - freezing point of solution
Let's find out m, molality (moles of solute in 1kg of solvent)
15 g / 58.45 g/mol = 0.257 moles of NaCl
NaCl(s) → Na⁺ (aq) + Cl⁻(aq)
i = 2 (Van't Hoff factor, numbers of ions dissolved)
m = mol /kg → 0.257 mol / 0.250kg = 1.03 m
Kf = Cryoscopic constant → 1.86 ºC/m (pure, for water)
0ºC - Tºf = 1.03m . 1.86ºC/m . 2
Tºf = -3.82ºC
Answer: The equilibrium constant is
Explanation:
Initial concentration of = 0.095 M
The given balanced equilibrium reaction is,
Initial conc. 0.095 M 0 M
At eqm. conc. (0.095-x) M (2x) M
Given : 2x = 0.0055
x = 0.00275
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :
Thus the equilibrium constant is
Answer:
3-bromo-4-ethyl-2,2,5-trimethyloctane.
Explanation:
You need to find the longest carbon chain. In this molecule, the longest carbon chain goes straight across and contains 8 carbons (oct-). There are no double or triple bonds so its octane.
Number the carbon chain from 1-8, starting at the end that is closest to a group. The left side is closest to a group so start from there.
On carbon 2, there are two methyl groups. On carbon 5, there is one methyl group. When naming, it will be 2,2,5-trimethyl.
On carbon 3, there is one bromine group. When naming, it will be 3-bromo.
On carbon 4, there is one ethyl group. When naming, it will be 4-ethyl.
Order the names of the groups alphabetically in front of octane.
The name of the compound is 3-bromo-4-ethyl-2,2,5-trimethyloctane.
Solubility product constants are values to describe the saturation of ionic compounds with low solubility. It is represented by Ksp and is the product of the concentration of the dissolved ions to the power of its stoichiometric coefficient. We do as follows:
Be(OH)2 = Be2+ + 2OH-
Ksp = [Be2+][OH-]^2
<span>6.92 × 10^-22 = x(2x)^2
x = 5.57x10^-8 mol /L
[Be(OH)2] = x = </span>5.57x10^-8 mol /L = 2.397x10^-6 g/L