Answer:
The final pressure of gas is 82.64 KNm⁻²
Explanation:
Given data:
Initial volume of gas = 180 cm³
Temperature of gas = 27°C
Initial pressure = 101 KNm⁻²
Final volume = 220 cm³
Final pressure = ?
Solution:
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
101 KNm⁻² × 180 cm³ = P₂
× 220 cm³
P₂ = 18180 KNm⁻². cm³/220 cm³
P₂ = 82.64 KNm⁻²
The final pressure of gas is 82.64 KNm⁻².
Answer:
Yes.
Explanation:
Hydrogen-filled balloons were widely used by the militaries during World War I (1914–1918). The main purpose of these hydrogen-filled balloons to detect movements of enemy troops and to provide direction to the artillery fire. Balloons were the targets of opposing aircraft because they knew the purpose of these balloons so they hit it whenever seen by the enemies so we can say that both sides used hydrogen-filled balloons as military observer to watch the enemy's movements.
The balanced chemical equation would be as follows:
<span>NH3+HCL->NH4CL
For this, we assume these gases are ideal gases so we can use the equation PV=nRT. We first calculate the number of moles of each reactants. We do as follows:
</span>PV=nRT
1.02 (4.21) = n (0.08206)(27+273.15)
n = 0.17 mol NH3 -------><span>consumed completely and therefore the limiting reactant</span>
PV=nRT
0.998 (5.35 L) = n (0.08206)(26+273.15)
n = 0.22 mol HCl
<span>what mass of NH4Cl(s) will be produced?
0.17 mol NH3 (1 mol NH4Cl / 1 mol NH3 ) = 0.17 mol NH3
which gas is the limiting reactant?
NH3 gas
which gas is present in excess?
HCl gas</span>
Answer:
1171.12 mL
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (milliliters)
V2 = final volume (milliliters)
T1 = initial temperature (Kelvin)
T2 = final temperature (Kelvin)
According to the information provided in this question:
P1 = 300 mmHg
P2 = 140 mmHg
V1 = 400 mL
V2 = ?
T1 = 0°C = 273K
T2 = 100°C = 100 + 273 = 373K
Using P1V1/T1 = P2V2/T2
300 × 400/273 = 140 × V2/373
120000/273 = 140V2/373
120000 × 373 = 273 × 140V2
44760000 = 38220V2
V2 = 44760000 ÷ 38220
V2 = 1171.115
The new volume is 1171.12 mL
