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: For a given mass and volume, how much physical space a material takes up, of an object or substance, the density remains constant at a given temperature and pressure. The equation for this relationship is ρ = m / V in which ρ (rho) is density, m is mass and V is volume, making the density unit kg/m3.
Explanation: Hope that helps
The fact that there are definite energy levels in the radiation spectra is known since light wavelength has been linked with the energy quantum (Einstein 1905 - Δ=ℎ.). These energy differences explain the reason why the spectrum of atoms is discontinuous (formed by spectral lines).
Bohr had the idea to link it to a change in the orbit radius of the electron around the nucleus (The idea that H atom could be formed by the association of one proton and one electron had been suggested a couple of years before by E. Rutherford).
This outstanding idea was perfected by a whole series of remarquable physicists in the years 1920–1930. It even continues to be refined to this day, where it forms the basis of atomic spectroscopy.
The plant makes 3.50 × 10⁷ bearings per day
<em>Pounds of bearings</em> = 1 da × (4355 lb bearings/5 da) = 871.0 lb bearings
<em>Grams of bearings</em> = 871.0 lb bearings × (453.6 g bearings/1 lb bearings)
= 3.951 × 10⁵ g bearings
<em>No. of bearings</em> = 3.951 × 10⁵ g bearings × (1 bearing/0.0113 g bearing)
= 3.50 × 10⁷ bearings
To dilute a solute in a solution, it is necessary to add a proper solvent for the reaction to occur, and adding more solvent will cause the solution to dilute even more, therefore the best answer will be letter B
