Assuming that the O2 gas acts like an ideal gas, we find
the following expression to be approximates of the behaviour of this gas:
<span>P V = n R T --->
1</span>
where,
P = pressure exerted by the gas
V = volume occupied
n = number of moles
R = universal gas constant
T = absolute temperature
Further, we assume that the number of moles and the
temperature are constant, hence reducing equation 1 into the form:
<span>P V = k --->
2</span>
where k is a constant. Therefore we can equate two
states:
P1 V1 = P2 V2
Since P1, V1 and V2 are given and we are to look for P2:
25 mL * 2 atm = 100 mL * P2
<span>P2 = 0.5 atm</span>
Answer:
It allows for easier division of the sister chromatids into two identical chromosomes, one for each cell, and helps to prevent mistakes. You just studied 2 terms!
Explanation:
Answer:
60 g/L is the final concentration of NaI solution .
Explanation:
Molarity of NaI solution before evaporation =
Volume of NaI solution before evaporation =
Molarity of NaI solution after evaporation =
Volume of NaI solution after evaporation =
( dilution)
Molar mass of NaI = 150 g/mol
Concentration of NaI after evaporation :
0.4 M × 150 g/mol = 60 g/L
60 g/L is the final concentration of NaI solution .
Answer:
The answer to your question is V2 = 0.203 kPa
Explanation:
Data
Volume 1 = V1 = 1l
Pressure 1 = P1 = 101.3 kPa
Volume 2= V2 = ?
Pressure 2 = 500 kPa
Process
- Use the Boyle's law to solve this problem
P1V1 = P2V2
-Solve for V2
V2 = P1V1/P2
-Substitution
V2 = (101.3 x 1) / 500
-Simplification
V2 = 101.3 / 500
-Result
V2 = 0.203 kPa
