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Answer:
0.259 kJ/mol ≅ 0.26 kJ/mol.
Explanation:
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 100.0 g).
c is the specific heat of water (c of ice = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).
<em>∵ Q = m.c.ΔT</em>
∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.
<em>∵ ΔH = Q/n</em>
n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.
∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.
Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
Solution:
At the equivalence point, moles NaOH = moles benzoic acid
HA + NaOH ==> NaA + H2O where HA is benzoic acid
At the equivalence point, all the benzoic acid ==> sodium benzoate
A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)
Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11
Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150
x^2 = 3.33x10^-12
x = 1.8x10^-6 = [OH-]
pOH = -log [OH-] = 5.74
pH = 14 - pOH = 8.26
Mass = 35.0 g
density = 1.036 g/cm³
volume = ?
therefore:
D = m / V
1.036 = 35.0 / V
V = 35.0 / 1.036
V = 33.784 cm³
answer D
