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Shkiper50 [21]
3 years ago
12

Suppose you used 0.5 M NaOH to titrate your vinegar sample instead of 0.1 M. What effect does the concentration of base added ha

ve on the reliability of the results of your titration?
Chemistry
1 answer:
laiz [17]3 years ago
4 0

Answer: the reliability will be worse

Explanation:

Suppose we used 0.5 M NaOH to titrate our vinegar sample instead of 0.1 M.

Now by using 0.5M instead of 0.1M we are increasing the concentration of NaOH,

We know that  the moles used = Volume x concnetration.

so for the same no of moles, if the concentration increases, the volume decreases.

Hence it will consume less NaOH.

now Since the volume decreases, the titration volume of less number will increase the % error.

Therefore the reliability will be worse.      

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Key terms: Reset ____________ anions are strong nucleophiles that open epoxide rings by an SN2 mechanism. ____________ tautomers
iogann1982 [59]

Answer:

Acetylide , Enol ,aldehydes, tautomers, alkynes , Hydroboration, Keto

Explanation:

Reset <u>Acetylide</u>  anions are strong nucleophiles that open epoxide rings by an SN2 mechanism. <u>Enol </u>tautomers have an O-H group bonded to a C=C. <u>aldehydes </u>are formed from terminal alkynes with the addition of water using BH3 then H2O2. <u>tautomers</u> are constitutional isomers that differ in the location of a double bond and a hydrogen and exist in an equilibrium with each other. <u>alkynes</u> are compounds that contain a carbon-carbon triple bond. <u>Hydroboration</u> of a terminal alkyne adds BH₂ to the less substituted, terminal carbon.<u> Keto</u> tautomers have a C=O and an additional C-H bond.

3 0
3 years ago
Dissolution of KOH, ΔHsoln:
swat32

Using Hess's law we found:

1) By <em>adding </em>reaction 10.2 with the <em>reverse </em>of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)   ΔH  (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the <em>total </em>change in enthalpy (ΔH).    

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq)   ΔHsoln    (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)     ΔHneut     (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from <u>differents changes</u> in various <em>reactions </em>can be calculated as the <u>sum</u> of all the <em>enthalpies</em> of all those <em>reactions</em>.      

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq)    (10.3)

We need to <em>add </em>reaction 10.2 to the <u>reverse</u> of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

<u>Canceling</u> the KOH(s) from both sides, we get <em>reaction 10.3</em>:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)    (10.3)

2) The change in enthalpy for <em>reaction 10.3</em> can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

\Delta H = \Delta H_{soln} + \Delta H_{neut}

The enthalpy of <em>reaction 10.1 </em>(ΔHsoln) changed its sign when we reversed reaction 10.1, so:

\Delta H = \Delta H_{neut} - \Delta H_{soln}

Therefore, the ΔHsoln must be <u>subtracted</u> from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

  • brainly.com/question/2082986?referrer=searchResults
  • brainly.com/question/1657608?referrer=searchResults  

I hope it helps you!

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