The answer for the following problem is mentioned below.
- <u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>
Explanation:
Given:
Initial pressure (
) = 290 kPa
Final pressure (
) = 104 kPa
Initial volume (
) = 18.9 ml
To find:
Final volume (
)
We know;
From the ideal gas equation;
P × V = n × R × T
where;
P represents the pressure of the gas
V represents the volume of gas
n represents the no of the moles
R represents the universal gas constant
T represents the temperature of the gas
So;
P × V = constant
P ∝ 
From the above equation;
represents the initial pressure of the gas
represents the final pressure of the gas
represents the initial volume of the gas
represents the final volume of the gas
Substituting the values of the above equation;
= 
= 52.7 ml
<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>
The average kinetic energy of an ideal gas is calculated as
KE_avg = 3/2 kT
where T is the temperature in Kelvin and k=R/N_A; R is the universal gas constant and N_A is the number of moles.
Thus, upon substitution we get
KE_avg = 3/2(8.314/1)(100+273)
KE_avg = 3/2(8.314)(373)
KE_avg = 4651.683
The average kinetic energy of 1 mole of a gas at 100 degree Celsius is 4651.683 J.
Answer:
See notes on LeChatlier's Principle I gave you yesterday.
Explanation:
Remember chemical see-saw => Removing Fe⁺³ makes the reactant side of the see-saw lighter causing the balance to tilt right then shift left to establish a new equilibrium with new concentration values. Such would result in a decrease in FeSCN⁺² concentration and increases in Fe⁺³ and SCN⁻ concentrations to replace the original amount of ppt'd Fe⁺³. => Answer Choice 'B' ... Also, see attached => Concentration effects on stability of chemical equilibrium .
Answer:
cell membrane
Explanation:
it controls what comes and goes out of a cell
To calculate the concentration of the base based on the titration, the concept used is the equal of number of equivalence of the acid used to that of the base. From this,
Na x Va = Nb x Vb
For HBr and KOH, molarity is equal to normality. Substituting the known values,
(0.75 M) x (22.6 mL - 0 mL) = Nb x (37.5 mL - 0.5 mL)
Nb = 0.46 N
Mb = 0.46 M
Thus, the concentration of the base is approximately 0.46 M.
