Answer:
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</h3>
Explanation:
First balance the chemical equation:
⇄ 
two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container
Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp
![K_p=[CO_2]](https://tex.z-dn.net/?f=K_p%3D%5BCO_2%5D)
![[CO_2]=p](https://tex.z-dn.net/?f=%5BCO_2%5D%3Dp)



Given mass of KNO₃=346g
Molar mass of KNO₃=(39.098)+(14)+(15.99*3)=101.068gmol⁻¹
Volume of Solution=750ml=0.75dm³
Molarity=(mass of solute/molar mass of solute)*(1/volume of sol. in dm³)
=(346/101.068)*(1/0.75)
=4.56 mol dm⁻³
Answer:
Which wind blows cool air inland during the day? Sea Breeze
Which wind blows cool air toward the sea at night? Land Breeze
Which winds blow steadily from specific directions and over long distances? Global Winds
Answer is: a lower freezing point has solution of K₂SO₄.
Change in freezing
point from pure solvent to solution: ΔT =i · Kf · b.<span>
Kf - molal freezing-point depression constant for water is 1.86°C/m.
b - molality, moles of solute per
kilogram of solvent.
i - </span>Van't
Hoff factor.<span>
b(K</span>₂SO₄<span>) = 0.35 m.
</span>b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.