<h3>
<u>Answer;</u></h3>
= 930.23 mL
<h3><u>Explanation</u>;</h3>
Using the combined gas law;
P1V1/T1 = P2V2/T2
Where; P1 = 600 kPa, V1 = 800 mL, and T1 = -25 +273 = 258 K, and
V2= ?, P2 = 1000 kPa, and T2 = 227 +273 = 500 K
Thus;
V2 = P1V1T2/T1P2
= (600 ×800 ×500) / (258 × 1000)
= 930.23 mL
When a 0. 1 m aqueous solution of hydrocyanic acid, HCN, reaches equilibrium, the ka for hydrocyanic acid is 3.969 x 10⁻¹⁰.
<h3>What is ka value?</h3>
It's the value of equilibrium constant for the dissociation of ions into a solution. The more the Ka value the more will be dissociation.
Ka = [H₃O⁺]² / [HCN] [H₃O⁺]
The pH is 5.20
-log [H₃O⁺] = 5.20
Putting antitlog both side.
The value will be 6.30 x 10⁻⁶
Ka = (6.30 x 10⁻⁶)² / 0.1 - 6.30 x 10⁻⁶
0.1 - 6.30 x 10⁻⁶ = 0.1
Ka = 3.969 x 10⁻¹⁰
Thus, the Ka value for hydrocyanic acid is 3.969 x 10⁻¹⁰.
To learn more about ka value, refer to the link:
brainly.com/question/2796803
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Voltage<span>, </span>also called<span> electromotive force, is a quantitative expression of the potential difference in charge between two points in an electrical field.
So ACTUALLY an "electromotive force", but of your answer choices.
D. Electrical Field Energy
</span>
Answer:
The heat of combustion for the unknown hydrocarbon is -29.87 kJ/mol
Explanation:
Heat capacity of the bomb calorimeter = C = 1.229 kJ/°C
Change in temperature of the bomb calorimeter = ΔT = 2.19°C
Heat absorbed by bomb calorimeter = Q
Moles of hydrocarbon burned in calorimeter = 0.0901 mol
Heat released on combustion = Q' = -Q = -2,692 kJ
The heat of combustion for the unknown hydrocarbon :
Answer: Least Common Multiplier of a6m, 0.003 moles 0.01elamos
Steps 6m, 0.003 moles
Compute an expression comprised of Factors that appear either in a6m or 0.003 moles
= 0.018elamos
Explanation:
