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3 years ago

6.

Chemistry

1 answer:

wlad13 [49]3 years ago

6 0

Answer:

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In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba(OH)2 was added to 60.0 mL of 0.600 M HCl. The reaction caused the temp

Zepler [3.9K]

Answer:

ΔH = 57.04 Kj/mole H₂O

Explanation:

60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)

=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)

=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)

=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat

ΔH = mcΔT/moles H₂O <==> Heat Transfer / mole H₂O

=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)

ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O

3 0

3 years ago

Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1

timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

7 0

3 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$