Where are the electrons found in Bohrs atomic atom

It is common for students to overshoot the endpoint, meaning they add too much NaOH(aq) from the buret, which causes the solutio

umka2103 [35]

Answer: the percentage of acetic acid will be low.

Explanation: The major aim during titration of acids and bases is to determine the endpoint , that is exact point where the acid in the beaker changes colour, (in this case, pink )with an additional drop from the burette containing the base, since it is usually difficult to mark the equivalence point that tells us when all the substrate in the beaker has been neutralized completely with the buretted substance.

Overshooting the end point is an error which can occur when the person involved in the the titration accidently goes beyond this endpoint by adding too much of the substance(base) from the burette into the beaker missing the exact endpoint.

This implies that the person has added too much of the burreted liquid, ie the base than required , making the acid in the beaker to continue to react resulting to a lower concentration of the acid (acetic acid) with excess base.(NaOH)

8 0

3 years ago

Can anyone explain enzymes in a simple way?

Sedaia [141]

Enzymes are biological catalysts which means they accelerate chemical reactions.
Hope this helps :)

4 0

2 years ago

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$