Answer:
Hydrogen bonding is a special type of dipole-dipole interaction that occurs between the lone pair of a highly electronegative atom (typically N, O, or F) and the hydrogen atom in a N–H, O–H, or F–H bond.
<span>1) </span><span>What is air pressure? (Answ</span>er: Atmospheric
pressure, sometimes also called barometric pressure, is the pressure within the
atmosphere of Earth.)
<span>2)
</span><span>What
is a barometer and what is it used to measure? <span>(Answer: A barometer is a scientific instrument used in meteorology to
measure atmospheric pressure. Pressure tendency can forecast short term changes in the
weather.)</span></span>
<span>3)
</span><span>How
does air pressure change? <span>(Answer: Millibar values used in
meteorology range from about 100 to 1050. At sea level, standard air pressure in millibars is
1013.2. Weather maps showing the pressure at
the surface are drawn using millibars. ... This change in pressure is
caused by changes in air density, and air density is related to
temperature.)</span></span>
<span>4) </span><span>Why is cooler, drier air related to High
Pressure? <span>(Answer: This is due to density differences
between the two air masses.
Since stronger high-pressure systems contain cooler or drier air, the air mass is denser and flows
towards areas that are warm or moist, which are in the vicinity of low pressure areas in advance of
their associated cold fronts.)</span></span>
<span>5)
</span><span>Why
is warm, moister air related to Low Pressure?
<span>(Answer: When air warms, its molecules scatter.
The air becomes
lighter and rises.)</span></span>
Answer : The concentration of HI (g) at equilibrium is, 0.643 M
Explanation :
The given chemical reaction is:
Initial conc. 0.10 0.10 0.50
At eqm. (0.10-x) (0.10-x) (0.50+2x)
As we are given:
The expression for equilibrium constant is:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
Now put all the given values in this expression, we get:
x = 0.0713 and x = 0.134
We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.0713
The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M
Thus, the concentration of HI (g) at equilibrium is, 0.643 M
Phosphorus is the correct answer.
