THe Answer is 105 J
hope this helps and please brainliest, I was first :)
The group is might be labeled as VIIB or VIIA.
Answer:
See the answer below
Explanation:
<em>The correct answer would be that the solute particles lower the solvent's vapor pressure, thus requiring a higher temperature to cause boiling.</em>
Dissolving a solute particle in a solvent leads to a decrease in the vapor pressure of the solvent above the resulting solution when compared to the pure solvent. The lower the vapor pressure of a liquid, the higher the temperature required for the liquid to boil and vice versa. Hence, a higher temperature would be needed to boil a solvent with dissolved solutes.
Answer:
- <u><em>It is positive when the bonds of the product store more energy than those of the reactants.</em></u>
Explanation:
The <em>standard enthalpy of formation</em>, <em>ΔHf</em>, is defined as the energy required to form 1 mole of a substance from its contituent elements under standard conditions of pressure and temperature.
Then, per defintion, when the elements are already at their standard states, there is not energy involved to form them from that very state; this is, the standard enthalpy of formation of the elements in their standard states is zero.
It is not zero for the compounds in its standard state, because energy should be released or absorbed to form the compounds from their consituent elements. Thus, the first choice is false.
When the bonds of the products store more energy than the those of the reactants, the difference is:
- ΔHf = ΔHf products - ΔHf reactants > 0, meaning that ΔHf is positive. Hence, the second statement is true.
Third is false because forming the compounds may require to use (absorb) or release (produce) energy, which means that ΔHf could be positive or negative.
Fourth statement is false, because the standard state of many elements is not liquid. For example, it is required to supply energy to iron to make it liquid. Thus, the enthalpy of formation of iron in liquid state is not zero.
The combination used in the preparation of a buffer with a pH around 8.8 accounts for ka = 7 * 10 -3 for h 3po4
The ph of the buffer can be shown as:
pH = pKa + log [Salt] /[ Acid ]
[Salt] /[ Acid ] = x
For h3po4 with ka= 7 × 10–3
8.8 = - log (7 × 10^–3) + log x
8.8 = 2.21 + log x
Thus, the value of log x is coming positive and therefore can be used for preparing buffer.
For h2po4- with ka= 8 × 10–8
8.8 = - log (8 × 10^–8) + log x
8.8 = 7.14 + log x
Thus, the value of log x is coming positive and therefore can be used for preparing buffer.
For hpo42– with ka= 5 × 10–13
8.8 = - log (5 × 10–13) + log x
8.8 = 12.31 + log x
Thus, the value of log x is coming negative and therefore can not be used for preparing buffer.
Hence, the correct answer is option A
Learn more about buffering systems here,
brainly.com/question/16556401
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