From the information given, the total volume of rubbing alcohol is 88.2 ml
68.6 % of this volume is isopropanol.
We will assume 88.2 ml represents 100% volume, so the volume of water will be 31.4 %
The volume of isopropanol is
68.6/100 x 88.2 → 0.686 × 88.2 = 60.505 ml
The volume of isopropanol is 60.5 ml.
Volume of water will be 88.20 - 60.5 = 27.7 ml
(27.7 / 88.2 × 100 = 31.4% )
Adding 60.5 ml of isopropanol to 27.7 ml of water to make up 88.2 ml will give 68.6 % v/v isopropanol to water solution.
N2 stands for nitrogen.
this compound also includes the exact mass.
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please note- I didn't quite understand the question fully.
-Diane
I believe the answer is D
Hey there!
Values Ka1 and Ka2 :
Ka1 => 8.0*10⁻⁵
Ka2 => 1.6*10⁻¹²
H2A + H2O -------> H3O⁺ + HA⁻
Ka2 is very less so I am not considering that dissociation.
Now Ka = 8.0*10⁻⁵ = [H3O⁺] [HA⁻] / [H2A]
lets concentration of H3O⁺ = X then above equation will be
8.0*10−5 = [x] [x] / [0.28 -x
8.0*10−5 = x² / [0.28 -x ]
x² + 8.0*10⁻⁵x - 2.24 * 10⁻⁵
solve the quardratic equation
X =0.004693 M
pH = -log[H⁺]
pH = - log [ 0.004693 ]
pH = 2.3285
Hope that helps!
Answer:
1400KJ/mol⁻¹
Explanation:
Amount of heat required can be found by:
Q = m × c × ΔT
<em>Where m is the mass, c is the specific heat capacity (4.2KJ for water) and ΔT is the change in temperature.</em>
Q = 24 × 4.2 × (23 - 9)
= 24 × 4.2 × 14
= 1411.2KJ/mol⁻¹
= <u>1400KJ/mol⁻¹</u> (to 2 significant figures)