When the dew point temperature and air temperature are equal, the air is said to be saturated. Dew point temperature is NEVER GREATER than the air temperature. Therefore, if the air cools, moisture must be removed from the air and this is accomplished through condensation.
It's classified as an acid
Answer:
v = 7.3 × 10⁶ m/s
Explanation:
Given data:
Velocity of electron = ?
Wavelength = 100 pm
Solution:
Formula:
λ = h/mv
λ = wavelength
h = planck's constant
m = mass
v = velocity
Now we will put the values in formula.
100 ×10⁻¹² m = 6.63 × 10⁻³⁴ j.s / 9.109 × 10⁻³¹ kg × v
v = 6.63 × 10⁻³⁴ kg.m²/s / 9.109 × 10⁻³¹ kg ×100 ×10⁻¹² m
v = 6.63 × 10⁻³⁴ m/s /910.9 × 10⁻⁴³
v = 0.0073 × 10⁹ m/s
v = 7.3 × 10⁶ m/s
<span>Pre-1982 definition of STP: 37 g/mol
Post-1982 definition of STP: 38 g/mol
This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using.
First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume.
Pre-1982: 2.1 / 22.414 = 0.093691443 moles
Post-1982: 2.1 / 22.71098 = 0.092466287 moles
Now determine the molar mass. Simply divide the mass by the moles. So
Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol
Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol
Finally, round to 2 significant figures. So
Pre-1982: 37 g/mol
Post-1982: 38 g/mol</span>
