Ok and bro???? Like what even lol
1. A
2. base
3. strong ones have more h30 weak ones have more OH
4. strong ones have more OH weak ones have more H30
5. <span>One of the main differences between acids and bases is that acids have a pH that is less than 7 and bases have a pH that is greater than 7. When dissolved in water, acids are substances that will cause the concentration of hydrogen ions (H+) to increase. Bases, when dissolved in water, will instead cause the number of hydroxide ions (OH-) to increase.6.Idk </span>
Answer:
B and D could be true
Explanation:
A volume of sodium hydroxide less than expected could occurs for two reasons:
The real concentration of sodium hydroxide was higher than expected or the amount of vinegar added was less than expected:
A. The sodium hydroxide solution had been allowed to stand exposed to the air for a long time prior to the titration. FALSE. A long expose to the air decreases concentration of the NaOH.
B. The volumetric flask used to prepare the diluted vinegar solution was rinsed with water prior to use. TRUE. You add a less amount of vinegar doing you require less amount of NaOH than expected.
C. The burette used to deliver the sodium hydroxide solution was rinsed with water prior to use. FALSE. Thus, you add a less amount of NaOH than expected. To explain the matter, you add more NaOH than expected.
D. The pipette used to deliver the vinegar solution was rinsed with water prior to use. TRUE. Again, you are adding a less amount of Vinegar than expected doing the necessary NaOH during titration less than expected
Answer:
ΔH = -20kJ
Explanation:
The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:
H₂(g) + S(g) → H₂S(g)
Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:
<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
The sum of -(1) + (2) + (3) gives:
<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>
<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>
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I hope it helps!
Answer:
Upper H superscript plus, plus upper O upper H superscript minus right arrow upper H subscript 2 upper O.
Explanation:
i just took the test , i hope this helps:)
