Answer:
The gas that Dr. Brightguy added was O₂
Explanation:
Ideal Gases Law to solve this:
P . V = n . R . T
Firstly, let's convert 736 Torr in atm
736 Torr is atmospheric pressure = 1 atm
20°C = 273 + 20 = 293 T°K
125 mL = 0.125L
0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K
(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n
5.20x10⁻³ mol = n
mass / mol = molar mass
0.1727 g / 5.20x10⁻³ mol = 33.2 g/m
This molar mass corresponds nearly to O₂
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Answer:
When the drill hits oil, some of the oil rises from the ground high into the air. This immediate release of oil is known as a "gusher." Once a reservoir has been located, pumps are used to extract the oil.
Explanation:
The answer is space between molecules decreases.
Hope this helps .>.
Answer:
Option a. 0.5 m/s
Explanation:
This graph shows a straight line, where "Y" axis would be "Position" and "X" graph would be "Time". The ecuation that would describe this straight line is Y= aX + 1 , where "a" is the slope or inclination for this graph, and would give us the speed of the object
How do we find the slope (and hence, the speed)?: if you notice this graph, you will check that:
-When X (Time) is zero, Y (Position) is 1
-When X (Time) is 2, Y (Position) is 2
With these 4 points, you can calculate the slope (which will call "m") for this graph with:
m = (Y2-Y1)/(X2-X1) so: Y2=2, Y1=1, X2=2, X1=0
Which gives us: m=1/2 (0.5), the slope or speed of the object: 0.5 m/s
