Answer:
Volume of container = 0.0012 m³ or 1.2 L or 1200 ml
Explanation:
Volume of butane = 5.0 ml
density = 0.60 g/ml
Room temperature (T) = 293.15 K
Normal pressure (P) = 1 atm = 101,325 pa
Ideal gas constant (R) = 8.3145 J/mole.K)
volume of container V = ?
Solution
To find out the volume of container we use ideal gas equation
PV = nRT
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
First we find out number of moles
<em>As Mass = density × volume</em>
mass of butane = 0.60 g/ml ×5.0 ml
mass of butane = 3 g
now find out number of moles (n)
n = mass / molar mass
n = 3 g / 58.12 g/mol
n = 0.05 mol
Now put all values in ideal gas equation
<em>PV = nRt</em>
<em>V = nRT/P</em>
V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa
V = 121.87 ÷ 101,325 pa
V = 0.0012 m³ OR 1.2 L OR 1200 ml
Liquid makes up most of the body
Answer:
D) both a and c are correct
Explanation:
The reaction rate is a measure of the speed of a chemical reaction. The factors that affects the rate of a chemical reaction are itemised below:
- Nature of the reactants
- Concentration of the reactants or pressure(if gaseous)
- Temperature
- Presence of catalyst
- Sunlight
Our concern here is temperature. Temperature affects a reaction considerably. Average kinetic energy is directly proportional to the temperature of the reacting particles. When the temperature of a reacting system is increase, the frequency of ordinary and effective collisions per unit time increases. A decrease in temperature implies that the number of collisions also decreases.
Answer:
6.68 X 10^-11
Explanation:
From the second Ka, you can calculate pKa = -log (Ka2) = 6.187
The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,
8.181 = (6.187 + pKa3) / 2
Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11
