Answer:
the volume occupied by 3.0 g of the gas is 16.8 L.
Explanation:
Given;
initial reacting mass of the helium gas, m₁ = 4.0 g
volume occupied by the helium gas, V = 22.4 L
pressure of the gas, P = 1 .0 atm
temperature of the gas, T = 0⁰C = 273 K
atomic mass of helium gas, M = 4.0 g/mol
initial number of moles of the gas is calculated as follows;

The number of moles of the gas when the reacting mass is 3.0 g;
m₂ = 3.0 g

The volume of the gas at 0.75 mol is determined using ideal gas law;
PV = nRT

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.
Is there an attachment because it depends on the element.
Answer:
The required volume of hexane is 0.66245 Liters.
Explanation:
Volume of octane = v=1.0 L=
Density of octane= d = 
Mass of octane ,m= 
Moles of octane =
Mole percentage of Hexane = 45%
Mole percentage of octane = 100% - 45% = 55%

Total moles = 11.212 mol
Moles of hexane :

Moles of hexane = 5.0454 mol
Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g
Density of the hexane,D = 
Volume of hexane = V

(1 cm^3= 0.001 L)
The required volume of hexane is 0.66245 Liters.
Yes the car is accelerating because it is changing direction as it goes around the corner
It is melting point process