Answer : The molar mass of unknown substance is, 39.7 g/mol
Explanation : Given,
Mass of unknown substance = 9.56 g
Volume of solution = 100.0 mL
Molarity = 2.41 M
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :

Now put all the given values in this formula, we get:


Therefore, the molar mass of unknown substance is, 39.7 g/mol
Answer : Option D) The atmospheric conditions vary as one changes latitude and altitude.
Explanation : The composition of the atmosphere varies according to the latitude and altitude because of the unequal heating of the earth surface at different latitudes and altitudes which results into atmospheric changes. It also creates different regions and zones.
Answer:
it is true.....................
Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL
Answer: The frequency of this light is 
Explanation:
To calculate the wavelength of light, we use the equation:

where,
= wavelength of the light =
c = speed of light = 
= frequency of light = ?

The frequency of this light is 