Answer:
0.062mol
Explanation:
Using ideal gas law as follows;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821Latm/molK)
T = temperature (K)
Based on the information provided;
P = 152 Kpa = 152/101 = 1.50atm
V = 0.97L
n = ?
T = 12°C = 12 + 273 = 285K
Using PV = nRT
n = PV/RT
n = (1.5 × 0.97) ÷ (0.0821 × 285)
n = 1.455 ÷ 23.39
n = 0.062mol
Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
volume of CL2
Solution:
Use the ideal gas law
PV = nRT
V = nRT/P
V = (2.1 moles Cl2) (0.08203 L - atm / mol - K) (273K) / (1 atm)
V = 47 L</span>
Answer:
704.6 g CO2
Explanation:
MM sucrose = 342.3 g/mol
MM CO2 = 44.01 g/mol
g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2
The increase in the boiling point of a solvent is a colligative property.
That means that the increase in the boling point will be related to the number of particles (molecules or ions) present in the solution.
The higher the number of particles (molecules or ions) the higher the increase in the boiling point.
All the aqueous solutions presented are electrolytes, i.e. the solutes are ionic compounds.
Then, you have to compare the number of ions that you have in each solution.
A) 1.0 M KCl ---> 1.0 M K+ + 1.0 MCl- = 2 moles of particles / liter
B) 1.0 M CaCl2 --> 1.0M Ca(2+) + 1.0M * 2 Cl (-) = 3 moles of particle / liter
C) 2.0M KCl ---> 2.0 M K+ + 2.0 M Cl- = 4 moles of particle / liter
D) 2.0 M CaCl2 ----> 2.0 M Ca (2+) + 2.0M * 2 Cl (-) = 6 moles of particle / liter.
Then, the solution 2.0M CaCl2(aq) has the highest increase in the boiling point.
Answer: option D) 2.0 M Ca Cl2(aq)
