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
Answer:
1. This reaction is <u>(A) Exothermic .</u>
2. When the temperature is decreased the equilibrium constant, K: <u>(A) Increases</u>
3.When the temperature is decreased the equilibrium concentration of Co2:<u> (A) Increases</u>
Explanation:
1. The pink color predominates at low temperatures, indicating that the commodity is preferred.
This is a reaction that is <u>exothermic.</u>
2. As the decrease in the temperature , the equilibrium constant , K ;
equilibrium constant = 
=![\frac{[CO^2^+][Cl^-^4]}{CoCl^2^-_4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO%5E2%5E%2B%5D%5BCl%5E-%5E4%5D%7D%7BCoCl%5E2%5E-_4%7D)
As the temperature drops, the concentration of
and 
rises, and K rises as well , thus it <u>increases </u>.
3. The equilibrium concentration of decreases as the temperature decreases:
When the temperature is lowered, the equilibrium shifts to the right , that is it <u>increases.</u>
The conversations need to solve this problem:
1 cal = 4.184 joules
1 Kcal= 1000 calories
1 kj= 1000 joules
or a more direct approach---->> 1 Kcal = 4.184 Kjoules
8.4 kcal (1000 calories/ 1 Kcal) x (4.184 joules/ 1 cal) x (1 Kj/ 1000 joules)= 35.1 Kj
or 8.4 kcal (4.184 Kj/ 1 kcal)= 35.1 Kj (same answer)
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
Answer:
True! Under most conditions, they are usually the same.
This is because to balance out the negativity or positivity of an atom, the opposite joins in.
Hope this helps.
