When Δ H > 0 that means this is an endothermic reaction, and we can assume heat as one of the reactants. SO when the temperature increases so the reaction will move rightward to decrease the reactants and increase the products and as the products increase it will make the Kw increase also as
Kw = [Products] /[reactants].
So our answer will be: the value of Kw will increase.
It will either be B or D, but I don't know about burial.
YES. Do I get brainliest now?
Answer:
The freezing point of the solution is -1.4°C
Explanation:
Freezing point decreases by the addition of a solute to the original solvent, <em>freezing point depression formula is:</em>
ΔT = kf×m×i
<em>Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.</em>
Molality of the solution is:
-moles CaBr₂ (Molar mass:
189.9g ₓ (1mol / 199.89g) = 0.95 moles
Molality is:
0.95 moles CaBr₂ / 3.75kg water = <em>0.253m</em>
Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:
CaBr₂(s) → Ca²⁺ + 2Br⁻
3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.
Replacing:
ΔT = kf×m×i
ΔT = 1.86°C/m×0.253m×3
ΔT = 1.4°C
The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,
<h3>The freezing point of the solution is -1.4°C</h3>
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Hello!
BeH₂ is a linear molecule, while CaH₂ is an angular molecule.
The difference between these two molecules is given by the number of electrons they have. Be is in the 2nd period of the Periodic table, and the ion Be⁺² doesn't have any free electron pairs when bonding to H. Ca is in the 4 period of the periodic table, meaning that it has more electrons, and the ion Ca⁺² has two free electron pairs when bonding to H that makes the molecule angular by pushing the bonds at an angle by sterical hindrance.
Have a nice day!
