From the information we have, this block of brown sugar has a volume of 8cm3
The mass of the block is 12. 9 grams.
We need to find out the density of the sugar.
For a solid material the formula for calculating density is given as:
Density = mass / volume
Therefore we simply fit in the above given values into this formula, so:
Density = 12.9 / 8
Density = 1.61
Therefore the density of the block of sugar is 1.61g/ml
Well it's definitely an alkaline solution.
Gibbous refers to the moon when it is more than half full but not completely full. This results in half a circle with a smaller arc on the opposite side. This makes a weird shaped “hump” on that side which is what the humpback means.
Answer:
The enthalpy of the nitrogen-nitrogen bond in N2H4 is 162.6 kJ
Explanation:
For the reaction: N2H4(g)+H2(g)→2NH3(g), the enthalpy change of reaction is
ΔH rxn = 2 ΔHºf NH3 - ΔHºf N2H4
but we also know that the ΔH rxn is calculated by accounting the sum of number of bonds formed and bonds broken as follows:
ΔH rxn = 6H (N-H) + 4 (N-H) + 2H (H-H)
where H is the bond enthalpy .When bonds are broken H is positive, and negative when formed, in the product there are 6 N-H bonds , and in the reactants 4 N-H and 1 H-H bonds).
Consulting an appropiate reference handbook or table the following values are used:
ΔHºf (NH3) = -46 kJ/mol
ΔHºf (N2H4) = 95.94 kJ/mol
(The enthalpy of fomation of hydrogen in its standard state is zero)
H (N-H) = 391 kJ
H (H-H) = 432 kJ
H (N-N) = ?
So plugging our values:
ΔH rxn = 2mol ( -46.0 kJ/mol) - 1mol(95.4 kJ/mol) = -187.40 kJ
-187.40 kJ = 6(-391 kJ) + 4 (391 kJ) + 432 + H(N-N)
-187.40 kJ = -350 kJ + H(N-N)
H(N-N) = 162.6 kJ
Answer:
a. 0.27 = Kc
b. 8.19×10⁻⁵ = Kp
Explanation:
The reaction is this: 3H₂(g) + N₂ (g) ⇄ 2NH₃ (g)
As we have the moles of each in the equilibrium and the volume is 1L, we assume the concentrations as molarity.
1.6981 mol/L → H₂
0.5660 mol/L → N₂
0.8679 mol/L → NH₃
Let's make the expression for Kc
Kc = [NH₃]² / [N₂] . [H₂]³
Kc = 0.8679² / 0.5660 . 1.6981³
Kc = 0.27
Let's calculate Kp, derivated from Kc
Kp = Kc . (RT)^Δn where:
Δn is the difference between final moles - initial moles. It is governed by stoichiometry. For this case 2 - (1+3) = -2
Δn it is always for gases
R is the Ideal gases constant
T is Absolute T°
Let's replace data → 0.27 . (0.082 . 700K)⁻² = Kp
8.19×10⁻⁵ = Kp
