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zloy xaker [14]
3 years ago
8

Difference between desirable changes and undesirable changes​

Chemistry
1 answer:
kkurt [141]3 years ago
3 0

Answer:

...n..m..mm .......mmnjjjjjjj ygttgyujjunjhh

Explanation:

ghyhhhhjn

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A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t
pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = pK_a=9.31

First we have to calculate the moles of HCN and NaCN.

\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole

and,

\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole

The balanced chemical reaction is:

                          HCN+NaOH\rightarrow NaCN+H_2O

Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

                        0.0655                       0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=9.31+\log (\frac{0.1359}{0.0655})

pH=9.63

Therefore, the pH of the solution is, 9.63

4 0
3 years ago
How are the three lines of defense the same? They all involve producing fever. They all involve fighting pathogens. They all inv
raketka [301]

Answer:

THEY ALL INVOLVE FIGHTING PATHOGENS

Explanation:

The immune system which is involved in defending the body against infections are diseases involves three lines of defense which are all involved in fighting against pathogens. Pathogens are invaders which when introduced into the body causes harm and therefore makes us sick. The body's first line of defense includes the physical barriers such as the skin, mucous membrane; chemical barriers such as tears, saliva, gastric acid in the stomach. These helps to keep the pathogens from entering the delicate parts of the body and once the pathogens find their way out of the reach of the first line of defense, the second line of defense is initiated. This includes inflammatory effects, swelling, redness, phagocytosis by neutrophils and macrophages. The third line of defense is the actions of lymphocytes which acts on invading microbes. The lymphocytes are of two types; the B and T cells. B cells produces antibodies which fight the antigens and T cells attack the infected cells of the body. There is also the memory cells which keeps information about the invading microbes for future attacks. This enables the body to respond swiftly when next the same type of pathogens attack.

6 0
3 years ago
Read 2 more answers
A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0
3 years ago
Which planets mass is nearly double that of mercury?
Nana76 [90]

Answer:

earth I think like that because earth is the second planet

4 0
3 years ago
A scientist prepared an aqueous solution of a 0.45 M weak acid. The pH of the solution was 2.72. What is the percentage ionizati
aleksandr82 [10.1K]

Answer:

0.42%

Explanation:

<em>∵ pH = - log[H⁺].</em>

2.72 = - log[H⁺]

∴ [H⁺] = 1.905 x 10⁻³.

<em>∵ [H⁺] = √Ka.C</em>

∴ [H⁺]² = Ka.C

∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.

<em>∵ Ka = α²C.</em>

Where, α is the degree of dissociation.

<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>

<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>

4 0
3 years ago
Read 2 more answers
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