1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
butalik [34]
3 years ago
11

Advantages and disadvantages of chemical​

Chemistry
1 answer:
Anit [1.1K]3 years ago
6 0

Answer:

Advantages :they are used for seeing chemical reactions .

they are used for doing chemical experiment .

they are used for making affective medicines .

Disadvantages :they harms to our environment.

they can make side effects in our body by affective medicines made by chemicals.

while doing different unknown experiments they make explosions which can also harm to us.

You might be interested in
6. A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCI. Calculate the pH
Vadim26 [7]

Answer:

a) pH = 9.14

b) pH = 8.98

c) pH = 8.79

Explanation:

In this case we have an acid base titration. We have a weak base in this case the pyridine (C₅H₅N) and a strong acid which is the HCl.

Now, we want the know the pH of the resulting solution when we add the following volume of acid: 0, 10 and 20.

To know this, we first need to know the equivalence point of this titration. This can be known using the following expression:

M₁V₁ = M₂V₂  (1)

Using this expression, we can calculate the volume of acid required to reach the equivalence point. Doing that we have:

M₁V₁ = M₂V₂

V₁ = M₂V₂ / M₁

V₁ = 0.125 * 25 / 0.1 = 31.25 mL

This means that the acid and base will reach the equivalence point at 31.25 mL of acid added. So, the volume of added acid of before, are all below this mark, so we can expect that the pH of this solution will be higher than 7, in other words, still basic.

To know the value of pH, we need to apply the following expression:

pH = 14 - pOH  (2)

the pOH can be calculated using this expression:

pOH = -log[OH⁻]  (3)

The [OH⁻] is a value that can be calculated when the pyridine is dissociated into it's ion. However, as this is a weak acid, the pyridine will not dissociate completely in solution, instead, only a part of it will be dissociated. Now, to know this, we need the Kb value of the pyridine.

The reported Kb value of the pyridine is 1.5x10⁻⁹ so, with this value we will do an ICE chart for each case, and then, calculate the value of the pH.

<u>a) 0 mL of acid added.</u>

In this case, the titration has not begun, so the concentration of the base will not be altered. Now, with the Kb value, let's write an ICE chart to calculate the [OH⁻], the pOH and then the pH:

       C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.125                                0             0

e)        -x                                   +x           +x

c)      0.125-x                              x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.125-x --> Kb is really small, so we can assume that x will be very small too, and 0.125-x can be neglected to only 0.125, and then:

1.5x10⁻⁹ = x² / 0.125

1.5x10⁻⁹ * 0.125 = x²

x = [OH⁻] = 1.37x10⁻⁵ M

Now, we can calculate the pOH:

pOH = -log(1.37x10⁻⁵) = 4.86

Finally the pH:

pH = 14 - 4.86

<h2>pH = 9.14</h2>

<u>b) 10 mL of acid added</u>

In this case the titration has begun so the acid starts to react with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.010) = 1x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 1x10⁻³ = 2.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 2.125x10⁻³ / (0.025 + 0.010) = 0.0607 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.0607                             0             0

e)        -x                                   +x           +x

c)      0.0607-x                           x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.0607-x --> 0.0607

1.5x10⁻⁹ = x² / 0.0607

1.5x10⁻⁹ * 0.0607 = x²

x = [OH⁻] = 9.54x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(9.54x10⁻⁶) = 5.02

Finally the pH:

pH = 14 - 5.02

<h2>pH = 8.98</h2>

<u>c) 20 mL of acid added:</u>

In this case the titration it's almost reaching the equivalence point and the acid is still reacting with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.020) = 2x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 2x10⁻³ = 1.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 1.125x10⁻³ / (0.025 + 0.020) = 0.025 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.025                                0             0

e)        -x                                   +x           +x

c)      0.025-x                             x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.025-x --> 0.025

1.5x10⁻⁹ = x² / 0.025

1.5x10⁻⁹ * 0.025 = x²

x = [OH⁻] = 6.12x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(6.12x10⁻⁶) = 5.21

Finally the pH:

pH = 14 - 5.21

<h2>pH = 8.79</h2>
5 0
3 years ago
Find the mass of a 60 ML volume of water if the density of water is 1 g/mL
valkas [14]

Answer:  60 grams

Explanation:  (60 ml)*(1g/ml) = 60g

3 0
3 years ago
Water changes from a liquid to a gas in the process of evaporation. <br> true or false
Umnica [9.8K]
This it true because in the triangle of the transformation from solid, liquid, and gas
:0<span />
4 0
3 years ago
Read 2 more answers
The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0°c is 4.48. what is the value of ka for hbro?
elena-14-01-66 [18.8K]

The solution would be like this for this specific problem:

Given:

 

pH of a 0.55 M hypobromous acid (HBrO) at 25.0 °C =  4.48

 

[H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-] <span>

Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>

 

To add, Hypobromous Acid does not require acid adjustment, which is necessary for chlorine-based product and is stable and effective in pH ranges of 5-9.<span>

</span>Hypobromous Acid combines with organic compounds to form a bromamine. Chlorine also combines with the same organic compounds to form a chloramine. <span>It is also one of the least expensive intervention antimicrobial compounds available.</span>

8 0
3 years ago
Read 2 more answers
Which of the following statements best describes what happens to water during vaporization?
ss7ja [257]

Answer: im pretty sure its D

Explanation:

8 0
3 years ago
Read 2 more answers
Other questions:
  • Mirrror metal and wood are the exaple
    11·1 answer
  • when mercury ii oxide is heated it decomposes into mercury and oxygen how many molecules of oxygen gas are produced if 12.5g of
    14·1 answer
  • Why Sncl2 is solid but SnCl4 is liquid at room temperature
    15·1 answer
  • Choose the most appropriate reagent(s) for producing the β-ketoester intermediate by alkylation of ethyl acetoacetate with 1-bro
    8·1 answer
  • What causes the appearance of lines in an emission spectrum?
    7·2 answers
  • Each particular chemical reaction cannot be expected to occur at identical rates.
    9·1 answer
  • . If one mole each of CH4, NH3, H2S, and CO2 is added to 1 liter of water in a flask (1 liter of water = 55.5 moles of H2O), how
    14·1 answer
  • What ions are produced when salt is dissolved in water
    12·2 answers
  • How can I identify when the reaction is in equilibrium
    13·1 answer
  • Name the sub atomic particles in part Z of a boron atom.Give the relative charges of these sub atomic
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!