Answer:
Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.
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Answer:
The warmer, lighter air rises, bringing cooler, heavier air to low altitudes.
Air at higher altitudes doesn't have as much air weighing down on it from above.
Explanation:
In short - air pressure is the result of the cumulative force that air molecules act on objects below them due to Earth's gravity. The higher the altitude, the less air molecules there are to act a force below them, and therefore, there's less air pressure at higher altitudes.
<span>First, write the net ionic equation for the unbalanced reaction. If you are given a word equation to balance, you'll need to be able to identify strong electrolytes, weak electrolytes and insoluble compounds. Strong electrolytes completely dissociate into their ions in water. Examples of strong electrolytes are strong acids, strong bases, and soluble salts. Weak electrolytes yield very few ions in solution, so they are represented by their molecular formula (not written as ions). Water, weak acids, and weak bases are examples of weak electrolytes. The pH of a solution can cause them to dissociate, but in those situations, you'll be presented an ionic equation, not a word problem. Insoluble compounds do not dissociate into ions, so they are represented by the molecular formula. A table is provided to help you determine whether or not a chemical is soluble, but it's a good idea to memorize the solubility rules.
</span><span><span>arate the net ionic equation into the two half-reactions. This means identifying and separating the reaction into an oxidation half-reaction and a reduction half-reaction. </span><span>For one of the half-reactions, balance the atoms except for O and H. You want the same number of atoms of each element on each side of the equation. </span><span>Repeat this with the other half-reaction. </span><span>Add H2O to balance the O atoms. Add H+ to balance the H atoms. The atoms (mass) should balance out now. </span><span>Now balance charge. Add e- (electrons) to one side of each half-reaction to balance charge. You may need to multiply the electrons the the two half-reactions to get the charge to balance out. It's fine to change coefficients as long as you change them on both sides of the equation. </span><span>Now, add the two half-reactions together. Inspect the final equation to make sure it is balanced. Electrons on both sides of the ionic equation must cancel out. </span><span>Double-check your work! Make sure there are equal numbers of each type of atom on both sides of the equation. Make sure the overall charge is the same on both sides of the ionic equation. </span><span>If the reaction takes place in a basic solution, add an equal number of OH- as you have H+ ions. Do this for both sides of the equation and combine H+ and OH- ions to form H2O. </span><span>Be sure to indicate the state of each species. Indicate solid with (s), liquid for (l), gas with (g), and aqueous solution with (aq). </span><span>Remember, a balanced net ionic equation only describes chemical species that participate in the reaction. Drop additional substances from the equation.ExampleThe net ionic equation for the reaction you get mixing 1 M HCl and 1 M NaOH is:H+(aq) + OH-(aq) → H2O(l)Even though sodium and chlorine exist in the reaction, the Cl- and Na+ ions are not written in the net ionic equation because they don't participate in the reaction.</span></span>
Answer:
The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.
Explanation:
Entropy :It is defined as amount of energy which is unable to do work or the measurement of randomness or disorderedness in a system.
Molar heat of molar vaporization of Trichlorofluoromethane = 24.8 kJ/mol
Temperature at which Trichlorofluoromethan boils , T= 296.95 K
The molar entropy of the evaporation of Trichlorofluoromethan :
The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.
The molarity of the acid given the data from the question is 0.30 M
<h3>Balanced equation </h3>
2HNO₃ + Ba(OH)₂ —> Ba(NO₃)₂ + 2H₂O
From the balanced equation above,
- The mole ratio of the acid, HNO₃ (nA) = 2
- The mole ratio of the base, Ba(NO₃)₂ (nB) = 1
<h3>How to determine the molarity of the acid</h3>
From the question given above, the following data were obtained:
- Volume of acid, HNO₃ (Va) = 39.7 mL
- Volume of base, Ba(NO₃)₂ (Vb) = 24 mL
- Molarity of base, Ba(NO₃)₂ (Cb) = 0.250 M
- Molarity of acid, HNO₃ (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 39.7) / (0.25 × 24) = 2
(Ma × 39.7) / 6 = 2
Cross multiply
Ma × 39.7 = 6 × 2
Ma × 39.7 = 12
Divide both side by 39.7
Ma = 12 / 39.7
Ma = 0.30 M
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