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.
The correct answer is option 2. A 0.8 M aqueous solution of NaCl has a higher boiling point and a lower freezing point than a 0.1 M aqueous solution of NaCl. This is explained by the colligative properties of solutions. For the two properties mentioned, the equation for the calculation of the depression and the elevation is expressed as: ΔT = -Km and <span>ΔT = Km, respectively. As we can see, concentration and the change in the property has a direct relationship.</span>
Answer:
It is necessary to use models to study sub- microscopic objects such as atoms and molecules because they are too small to be seen.
Answer:
Rate constant = 0.0237 M-1 s-1, Order = Second order
Explanation:
In this problem, it can be observed that as the concentration decreases, the half life increases. This means the concentration of the reactant is inversely proportional to the half life.
The order of reaction that exhibit this relationship is the second order of reaction.
In the second order of reaction, the relationship between rate constant and half life is given as;
t1/2 = 1 / k[A]o
Where;
k = rate constant
[A]o = Initial concentration
k = 1 / t1/2 [A]
Uisng the following values;
k = ?
t1/2 = 113
[A]o = 0.372M
k = 1 / (113)(0.372)
k = 1 / 42.036 = 0.0237 M-1 s-1
The answer is “B” because I just answered it
