Gravity is 9.8 meters per second. It's a 9.2 mile fall. so just divide the two and you get 0.9388 seconds.
Let's assume ideal gas behavior for simplicity. We could use the equation below:
PV=nRT
Solve for n or the number of moles.
n = PV/RT = (3×10⁻³ atm)(1 L)/(0.0821 L·atm/mol·K)(250 K)
n = 1.462×10⁻⁴ moles ozone
For every 1 mole of any substance, Avogadro stipulated that there is an equivalent of 6.022×10²³ molecules. Thus,
# of ozone molecules = 1.462×10⁻⁴ mol * 6.022×10²³ molecules/1 mol
<em># of ozone molecules = 8.8×10¹⁹</em>
Density is equal to the ratio of mass to the volume.
The mathematical expression is given as:
Arrange the above expression in terms of volume, i.e.
Density of osmium is equal to =
Mass of osmium = 4.80 kg
First, convert the mass in kg to g:
1 kg = 1000 g
Therefore, mass in g = = 4800 g
Put the values,
= .
Hence, volume occupied by osmium =.
Answer: 0.082 atm L k^-1 mole^-1
Explanation:
Given that:
Volume of gas (V) = 62.0 L
Temperature of gas (T) = 100°C
Convert 100°C to Kelvin by adding 273
(100°C + 273 = 373K)
Pressure of gas (P) = 250 kPa
[Convert pressure in kilopascal to atmospheres
101.325 kPa = 1 atm
250 kPa = 250/101.325 = 2.467 atm]
Number of moles (n) = 5.00 moles
Gas constant (R) = ?
To get the gas constant, apply the formula for ideal gas equation
pV = nRT
2.467 atm x 62.0L = 5.00 moles x R x 373K
152.954 atm•L = 1865 K•mole x R
To get the value of R, divide both sides by 1865 K•mole
152.954 atm•L / 1865 K•mole = 1865 K•mole•R / 1865 K•mole
0.082 atm•L•K^-1•mole^-1 = R
Thus, the value of gas constant is 0.082 atm L k^-1 mole^-1