Answer:

Explanation:
Given:
volume of air in the room, 
temperature of the room, 
<u>Saturation water vapor pressure at any temperature T K is given as:</u>
<u />
<u />
putting T=298 K we have

<u>The no. of moles of water molecules that this volume of air can hold is:</u>
Using Ideal gas law,



is the maximum capacity of the given volume of air to hold the moisture.
Currently we have 80% of n, so the mass of 20% of n:

where;
M= molecular mass of water

is the mass of water that can vaporize further.
<span>A red ladybug appears red in white light,red in red light, and black in blue light.</span>
Answer:
34.8 and 55.2º
Explanation:
This is a projectile launching exercise, as we are told that the range of the arrow must be equal to its range and = 31 m let's use the equation
The scope equation is
R = v₀² sin 2θ /g
sin 2 θ = R g / v₀²
sin 2 θ = 31 9.8 / 18²
2 θ = sin⁻¹ 0.93765
θ = 34.8º
At the launch of projectiles we have two complementary angles with the same range in this case 34.8 and (90-34.8) = 55.2º
Answer:
Vibrate
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