<span>The </span>standard enthalpy of formation<span> <span>is defined as the change in </span></span>enthalpy<span> <span>when one mole of a substance in the </span></span>standard<span> <span>state (1 atm of pressure and temperature of 298.15
K) is </span></span>formed<span> <span>from its
pure elements under the same conditions.</span></span>
Answer: D
Explanation: Keeping the pressure constant and increasing the temperature.
For the answer to the question above, I can't help you directly because I don't have a calculator right now. But I'll show you how to solve this.
<span>use the freezing point depression formula for this one: delta T = i * m * K where K is a constant, m is the molality (mol solute/kg solvent), and i is the van'hoff factor the van hoff factor is the number of ions that your salt dissociates into. Since it's an ALKALI flouride salt, how many ions? k is just a constant, you get it from a table in your textbook somewhere So you have everything to solve for the molality of the solution, once you did that, multiplying it by the mass of water to find the mols of the salt. Take the mass of the salt and divide by this mols to figure out the molar mass, and then compare it with the periodic table to identify the salt.
</span>
<u>Mole solute</u> x mass of Water = Mol solute<u>
</u>kg Solvent
then
Mass of solute x <u> 1 </u> = molar mass
mole of solute
The original potassium atom then becomes a potassium cation with formula k+ the potassium atom donates one of its electrons, to be specific the only the electron in its valence shell to another more electronegative atom.
