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MAVERICK [17]
4 years ago
15

In this experiment, you used phenolphthalein to monitor the neutralization reactions. describe the chemical differences between

an equivalence point and an endpoint in titration and use this to explain whether phenolphthalein could be used as an indicator in any acid/base neutralization.
Chemistry
1 answer:
viva [34]4 years ago
6 0
The endpoint in tiration is the point where an indicatior's halfway thru its color change. Equivalence point is where moles/stoichiometry of the system is satisfied (moles of reactants are equal to each other).

Titrating a strong acid with a strong base results in a salt that is neutral. Phenolphthalein changes color in the range <span>8.3 – 10. It is very easy to spot the change as it is colorless in acidic (< 8.3) and pink in basic (> 10).

pH will rapidly change near titration equivalence point. </span><span>Only one drop of the titrant causes this large change, the color change of phenolphthalein does not occur on the equivalence point, but IT IS within about 1 drop. <span>It would be considered an "acceptable uncertainty" in using titration to determine concentration by volumetric measurement.</span></span>

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Calculate the mass (g) of each product formed when 43.82 g of diborane (B₂H₆)
guajiro [1.7K]

Stoichiometry:

The area of study in chemistry concerned with the quantities of chemical species produced or required for a given chemical reaction is stoichiometry. The stoichiometric coefficients of the balanced chemical equation are often used to relate the amounts of substances to one another.

Evaluation :

The balanced chemical equation for the reaction is given as follows:

B_{2} H_{6} + 6H_{2} O  → 2H_{3} BO_{3}  + 3H_{2}

By using the molar mass of the given chemical species, along with the stoichiometric coefficients of the balanced chemical equation, we determine the mass of each product as:

43.82 g B_{2} H_{6} ×\frac{1 Mol B_{2} H_{6} }{27.66 g} ×\frac{3molH_{2} }{1 molB_{2}H_{6}  } ×\frac{2g}{1 mol H_{2} } = 9.51 g H_{2}

43.82 g B_{2} H_{6} × \frac{1 mol B_{2} H_{6} }{27.66 g}    ×   \frac{1 mol H_{3}BO_{3}  }{1 mol B_{2} H_{6} } × \frac{61.8g}{1 mol H_{3} BO_{3} }  = 196 gH_{3} BO_{3}

Learn more about chemical reaction :

brainly.com/question/1893305

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6 0
2 years ago
Consider the following multistep reaction: A+B→AB(slow) A+AB→A2B(fast)−−−−−−−−−−−−−−−−− 2A+B→A2B(overall) Based on this mechanis
Grace [21]

Answer:

r = k × [A] × [B]

Explanation:

To determine the rate law, we simply use the slow step reaction equation. The slow step is the rate determining step in the reaction.

A+B→AB

And as we know, the rate of the reaction is proportional directly to the product of the concentration of the reactants which concentration is changing over the course of the reaction.

r = k × [A] × [B]

Where r = rate of reaction

k = reaction rate constant

[A] = Concentration of molecule A

[B] = Concentration of molecule B

8 0
3 years ago
List four observations that suggest chemical change is occurring
allsm [11]

Answer:

Examples of Chemical Changes

Burning wood.

Souring milk.

Mixing acid and base.

Digesting food.

Cooking an egg.

Heating sugar to form caramel.

Baking a cake.

Rusting of iron.

8 0
3 years ago
Did I get This right I’m confused
Paul [167]
Hey bud !
You were close congrats !
the only problem i saw was the units part at the end.
remember volume is 3 dimensional
length width and height
so it is m^3
3 0
4 years ago
A covalent chemical bond is one in which
vitfil [10]
I think the answer is c
4 0
3 years ago
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