Example of a substitution method How to solve problem using substitution method By part rule with examples integrate sec^2x ln(tanx +2)dx

Example of a substitution method
How to solve problem using substitution method
By part rule with examples

End point , Molarity ,pH value problems

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problem -1

The titration of 10.00 mL of a diprotic acid solution of unknown concentration requires 21.37 mL of a 0.1432 M NaOH solution. What is the concentration of the diprotic acid solution?

problem -2  

The titration of 10.00 mL of a diprotic acid solution of unknown concentration requires 21.37 mL of a 0.1432 M NaOH solution. What is the concentration of the diprotic acid solution?

problem -3 

A 0.2602 g sample of an unknown monoprotic acid requires 12.23 mL of 0.1298 M NaOH solution to reach the end point. what is the molecular mass of the acid?

problem -4 

 What is the final pH after 1 drop (0.05 mL) of 6 M HCI is added to 1.0 L of freshly prepared pure water that was originally at a pH of 7.0. Is there a significant pH change?

problem - 5 

What is the molarity of pure water at 20 C?

problem 6 

Density table of water at one atm pressure 

 

How many of the following are pure compounds? sodium, sugar, oxygen, air, iron

How many of the following are pure compounds? sodium, sugar, oxygen, air, iron

Answer :-

Sodium is a element 

Sugar is a compound 

Oxygen is also a element 

Air is a mixture 

Iron is element 

so answer of this problem is sugar 

Which of the following processes require(s) chemical methods? 
a)Separating a homogeneous mixture into pure substances.
b)Separating a heterogeneous mixture into pure substances. 
c)Distilling a saltwater mixture. 
d)Breaking a compound into its constituent elements. 
At least two of the above (a-d) require chemical methods

Answer :-

a)Separating a homogeneous mixture into pure substances. Its can involve both physical and chemical method for the separation

b)Separating a heterogeneous mixture into pure substances. IT involve physical method only 

c)Distilling a saltwater mixture it is a physical method

d)Breaking a compound into its constituent elements.  this is a chemical reaction

so Answer is a and d

Which is an example of a homogeneous mixture? 
a)vodka
b)oily water 
c)soil (dust) 
d)sodium chloride aluminum


in this problem Vodka is homogeneous mixture because  in out of four option,only vodka has a uniform composition throughout 

Assume the equilibrium constant for the reaction of a particular alcohol with a carboxylic acid is 7.87. What is the calculated yield of the ester?

 Assume the equilibrium constant for the reaction of a particular alcohol with a carboxylic acid is 7.87. What is the calculated yield of the ester?
Assume the equilibrium constant for the reaction of a particular alcohol with a carboxylic acid is 7.87. What is the calculated yield of the ester?

If you want answer of this problem please mail me on nirapendra.singh@gmail.com 

A calorimeter contains 35.0mL of water at 15.0 degrees celcius . When 1.20g of X (a substance with a molar mass of 51.0g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)--->X(aq) and the temperature of the solution increases to 29.0 degrees celcius . Calculate the enthalpy change,(deltaH), for this reaction per mole of X. Assume that the specific heat and density of the resulting solution are equal to those of water [4.18 (J/g(degree celcius)) and 1.00g/mL ] and that no heat is lost to the calorimeter itself, nor to the surroundings.

Prob -1  
A calorimeter contains 35.0mL of water at 15.0 degrees celcius . When 1.20g of X (a substance with a molar mass of 51.0g/mol ) is added, it dissolves via the reaction
X(s)+H2O(l)--->X(aq)
and the temperature of the solution increases to 29.0 degrees celcius .
Calculate the enthalpy change,(deltaH), for this reaction per mole of X.
Assume that the specific heat and density of the resulting solution are equal to those of water [4.18 (J/g(degree celcius)) and 1.00g/mL ] and that no heat is lost to the calorimeter itself, nor to the surroundings.


In this problem heat is consumed in increasing the temperature of water


heat  Q = M*C* delta T

M = mass of water = 35.0 ml * 1 gram / 1 ml  =  35 gram   (density of water is 1gram per cm ^ 3 or 1.00g/mL  )
 
delta T = final temperature - initial  temperature
          =  29 - 15  =  14 `C
 
specific heat capacity C = 4.18 (J/g(degree celcius)
 put these value in equation we get heat  Q = M*C* delta T
Heat Q = 35 * 4.18*14   
           = 2048.2j


When 1.20g of X (a substance with a molar mass of 51.0g/mol ) is added
number of moles of substance X =  Mass / molar mass 
                                               = 1.20/51.0   
                                               = 0.0235 mol

Molar heat capacity = heat required / number of moles
                              = 2048.2 j / 0.035 mol
                              = 87048.5 jules                          (1000 j = 1 kj )
convert in kilojoules
                             = 87048.5/ 1000 kj
                             =87.05 kj                            

         Answer = 87.05 kj

Problem - 2  This one problem has me confused
I need 60.0 in scientific notation
is it 6.0x10^1 ??

your answer is correct 

in scientific notation we always place decimal point after one digit from the left

example

60000000   = 6.0 x10^7
0.00006 = 6.0 x10^ -5

and  60.0  =  6.0x10^1   =  6.0x10  

Find the four smallest positive numbers 0 such that tan 0=-1 [Write your answers in ascending order.]

Find the four smallest positive numbers 0 such that tan 0=-1 [Write your answers in ascending order.]

Given the following delta H values H2+1/2O2--->H2O delta H =-285.8 H2O2---->H2+O2 delta H = 187.6 Calculate delta H rxn for the following reaction H2O2--->H2O + 1/2O2

14.77 mL of 0.4926 M NaOH is used to titrate 10.00 mL of vinegar to the end point. What is the molarity of acetic acid present in the vinegar solution? Hint: How many moles of acetic acid was titrated?

Hi Nyuyen this is your answer

trigonometric values 0 to 90

Double angle formulas of sinA, Cos A and tan A

Double angle formulas of  sin 2A , Cos 2A and tan2A 

             sin(2A)   = 2sinAcosA

             cos(2A)  = 2cos^2A- sin^2 A 

             cos(2A)  = 1- 2sin^2A

             cos(2A)  =  2cos^2A - 1 

             tan (2A)  =  2tanA / (1- tan^2A)

Double angle formula for sinA.cosA and tan A

g(t)=1-2t-3t^2, using the limit definition of derivative, find the tangent line to g(t) at t=-2

convert the following into basic units: 28.7pm 15.15pm 25365mg (with explanation )

It takes 46.0J to raise the temperature of an 8.00g piece of unknown metal from 13.0\rm ^\circ C to 24.3^\circ{C}. What is the specific heat for the metal? Express your answer numerically, in \rm J/g \cdot {}^\circ C

It takes 46.0J to raise the temperature of an 8.00g piece of unknown metal from 13.0\rm ^\circ C to 24.3^\circ{C}. What is the specific heat for the metal?
Express your answer numerically, in \rm J/g \cdot {}^\circ C

If the speed of light is 3.0 ×10^8 m s-1, calculate the distance covered by light in 2.00 ns

Q-22 If the speed of light is 3.0 ×10^8 m s-1, calculate the distance covered by light in 2.00 ns.
Answer
According to the question: 

1ns  = 10^-9 s (conversion )

Time taken to cover the distance = 2.00 ns 

                                                 = 2.00 × 10^–9 s
            Speed of light                = 3.0 × 10^8 ms–1
Distance= Speed of light × Time taken  (formula )
= (3.0 × 10^8 ms –1) (2.00 × 10^–9 s)
= 6.00 × 10^–1 m
= 0.600 m

The following data are obtained when dinitrogen and dioxygen react together to form different compounds: Mass of dinitrogen Mass of dioxygen (i) 14 g 16 g (ii) 14 g 32 g (iii) 28 g 32 g (iv) 28 g 80 g

Q- 21 The following data are obtained when dinitrogen and dioxygen react together to form
different compounds:
Mass of dinitrogen Mass of dioxygen
(i) 14 g 16 g
(ii) 14 g 32 g
(iii) 28 g 32 g
(iv) 28 g 80 g
Answer :- 
(a) Which law of chemical combination is obeyed by the above experimental data?Give
its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = ...................... mm = ...................... pm
(ii) 1 mg = ...................... kg = ...................... ng
(iii) 1 mL = ...................... L = ...................... dm3

Answer :-

A) Fixing the mass of dinitrogen to 28g , mass of dioxygen combined with 32,64,32, and 80 g in the given four oxides . This is the ratio of 1:2:1:5 which is a simple whole number ratio . Hence the given data follow the law of multiple proportions

Law of Multiple Proportions This law was proposed by Dalton in 1803.According to this law, if two elements cancombine to form more than one compound, themasses of one element that combine with afixed mass of the other element, are in theratio of small whole numbers.For example, hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.
Hydrogen + Oxygen → Water
2g                 16g           18g
Hydrogen + Oxygen → Hydrogen Peroxide
2g               32g                 34g
Here, the masses of oxygen (i.e. 16 g and 32 g)which combine with a fixed mass of hydrogen (2g) bear a simple ratio, i.e. 16:32 or 1: 2. .

B)