Not elementary school work
<u>Given:</u>
Moles of He = 15
Moles of N2 = 5
Pressure (P) = 1.01 atm
Temperature (T) = 300 K
<u>To determine:</u>
The volume (V) of the balloon
<u>Explanation:</u>
From the ideal gas law:
PV = nRT
where P = pressure of the gas
V = volume
n = number of moles of the gas
T = temperature
R = gas constant = 0.0821 L-atm/mol-K
In this case we have:-
n(total) = 15 + 5 = 20 moles
P = 1.01 atm and T = 300K
V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L
Ans: Volume of the balloon is around 488 L
Answer:
58.9mL
Explanation:
Given parameters:
Initial volume = 34.3mL = 0.0343dm³
Initial concentration = 1.72mM = 1.72 x 10⁻³moldm⁻³
Final concentration = 1.00mM = 1 x 10⁻³ moldm⁻³
Unknown:
Final volume =?
Solution:
Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.
Therefore;
C₁V₁ = C₂V₂
where C and V are concentration and 1 and 2 are initial and final states.
now input the variables;
1.72 x 10⁻³ x 0.0343 = 1 x 10⁻³ x V₂
V₂ = 0.0589dm³ = 58.9mL
The answer is a. plenty of water
The volume of the gas that occupy at STP is 165. 28 cm^3
calculation
by use of combined gas law that is P1V1/T1=P2V2/T2, where
P1=84.6 kpa
T1=23.5 +273=296.5 K
V1=215 cm^3
At STP T= 273 K and P= 101.325 Kpa
therefore p2 = 101.325 Kpa and T2 = 272 K V2=?
by making V2 the subject of the formula V2 =T2P1V1/P2T1
V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3
