When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

Given the specific heat capacity of the solution (c) is 4.184 J/g・°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
Learn more: brainly.com/question/4400908
Answer: 19.25 gallons
Explanation: 1 ml = 0.0011 quart
Given: 4 quarts = 1 gallon
Thus if 1 ml is equal to 0.0011 quart
70000 ml is equal to =
Now if 4 quarts is equal to 1 gallon.
77 quarts is is equal to=
Answer:He 1+
Explanation:
Both Hydrogen atom and helium ion are one electron species. Hence we expect the spectrum of the helium ion to closely resemble that of hydrogen atom also containing one valence electron.
Answer:
Answers are in the explanation.
Explanation:
<em>Given concentrations are:</em>
- <em>SO₂ = 0.20M O₂ = 0.60M SO₃ = 0.60M</em>
- <em>SO₂ = 0.14M O₂ = 0.10M SO₃ = 0.40M </em>
- <em>And SO₂ = 0.90M O₂ = 0.50M SO₃ = 0.10M</em>
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In the reaction:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
Kc is defined as:
Kc = 15 = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are equilbrium concentrations.</em>
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Also, you can define Q (Reaction quotient) as:
Q = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are ACTUAL concentrations.</em>
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If Q > Kc, the reaction will shift to the left until Q = Kc;
If Q < Kc, the reaction will shift to the right until Q = Kc
If Q = Kc, there is no net reaction because reaction would be en equilibrium.
Replacing with given concentrations:
- Q = [0.60M]² / [0.60M] [0.20M]² = 15; Q = Kc → No net reaction
- Q = [0.40M]² / [0.10M] [0.14M]² = 82; Q > Kc, → Reaction will shift to the left
- Q = [0.10M]² / [0.50M] [0.90M]² = 0.015; Q < Kc → Reaction will shift to the right
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