6.022*10^23 atoms/mole of reactant
(this is chemistry not physics)
Answer:
Both
Explanation:
The lithosphere is part of both the crust and the mantle.
It is the surface layer of the earth and also the most rigid layer. It is formed by the crust and the outermost part of the mantle. It is divided into two types: continental lithosphere and oceanic lithosphere.
The oceanic lithosphere has an approximate thickness of 50 - 100km, and the continental olithosphere of 40 - 200km.
Answer:
(a). 14.4 lbf/in^2.
(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Explanation:
So, from the question above we are given the following parameters which are going to help us in solving this particular Question;
=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"
=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."
With these knowledge, let us delve right into the solution;
(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].
The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.
(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.
Length of mercury = 27.8 in.
AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Answer:
r = 0.0548 m
Explanation:
Given that,
Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.
We need to find the radius of their circular path. The formula for the radius of path is given by :
m is mass of Singly charged uranium-238 ion, 
q is charge
So,
So, the radius of their circular path is equal to 0.0548 m.
