A card is selected from a standard deck of 52 playing cards. A standard deck of cards has 12 face cards and four Aces (Aces are not face cards). Find the probability of selecting · a five given the card is a not a club. · a heart given the card is red. · a face card, given that the card is black. Show step by step work. Give all solutions exactly in reduced fraction form

A card is selected from a standard deck of 52 playing cards. A standard deck of cards has 12 face cards and four Aces (Aces are not face cards). 

Find the probability of selecting · 

1.  a five given the card is a not a club. 
2.  a heart given the card is red. 
3.  a face card, given that the card is black.

 Show step by step work. Give all solutions exactly in reduced fraction form

Answer : -

1) there are total 4 five cards but one of them is five of club 

so number of 5 which is not club are 3   

 and there are total 13 clubs card 

       so non clubs card will 52- 13 will  39

formula of probability = favorable out comes / total out comes 

 so probability of  a five given the card is a not a club.                                                                    = 3/39

2) there are total 13 heart card 

    and total red cards are 26 

so that probability of  card is red  

                                            = 13/26 

                                            = 1/2

3) there are total 26 black card 

  there will 3 face card of spades (♠) and 

                3 face card of  clubs (♣)

 so total card will be 6 of spades and clubs

probability of a face card, given that the card is black

                                         = favorable out comes / total out comes 

                                         =6/26 

                                         = 3/13

A card is selected from a standard deck of 52 playing cards. A standard deck of cards has 12 face cards and four Aces (Aces are not face cards). Find the probability of selecting · a five given the card is a not a club. · a heart given the card is red. · a face card, given that the card is black. Show step by step work. Give all solutions exactly in reduced fraction form

St andrews golf course in st andrews scotland is one of the oldest courses in the world. it is an 18-hole course that consists of par-3 holes, par-4 holes and par-5 holes. a golfer who shoots par at the old course at st andrews has a total of 72 strokes for the entire course. there are seven times as many par-4 holes as par- 5 holes,and the sum of the numbers of par-3 and par-5 holes is four.find the numbers of par-3, par-4 and par-5 holes in the course?

St andrews golf course in st andrews scotland is one of the oldest courses in the world. it is an 18-hole course that consists of par-3 holes, par-4 holes and par-5 holes. a golfer who shoots par at the old course at st andrews has a total of 72 strokes for the entire course. there are seven times as many par-4 holes as par- 5 holes,and the sum of the numbers of par-3 and par-5 holes is four.find the numbers of par-3, par-4 and par-5 holes in the course?
                let x represent the par 3   

                    y represent the pr 4 

                    z represent the par 5 

       18-hole course that consists of par-3 holes, par-4 holes and par-5 holes.

                                            so     x + y + z = 18    ..............(1)

                          seven times as many par -4 holes as par -5 holes

                                           so                 y = 7z      .............(2)

sum of the number of par-3 and par 5 holes is four 

                                           so          x + z  = 4 

             subtract   z both side we get 

                                                              x  = 4- z                  ................(3)

 put the value of  x and y in equation first we get

                                              4-z + 7z + z = 18 

                                                      4 + 7z  = 18 

                                                              7z =  18- 4 

                                                                z = 14/7

                                                                z = 2   

put the value of z in equation second and third we get 

                                                              y = 7z  

                                                                 = 7* 2 

                                                                 = 14 

                                                     and    x  = 4-z 

                                                                   =4- 2 

                                                                   = 2

 Answer is   x= 2 , y = 14 and z = 2

Problem  St andrews golf course in st andrews scotland is one of the oldest courses in the world. it is an 18-hole course that consists of par-3 holes, par-4 holes and par-5 holes. a golfer who shoots par at the old course at st andrews has a total of 72 strokes for the entire course. there are seven times as many par-4 holes as par- 5 holes,and the sum of the numbers of par-3 and par-5 holes is four.find the numbers of par-3, par-4 and par-5 holes in the course

 

Determine specific heat in J/g times degree celcius . Mass of substance 29.2g, initial temp 34.4degrees celcius, final temp 68.1degrees celcius, heat applied 436.9 j, what is the formula of hear capacity ?

Q- 1 What is the heat capacity ?

Answer :- Amount of heat which required to raise the temperature of one gram substance by one degree Centigrade is called  heat capacity of that substance.

example:-

heat capacity of water is 4.18 j /gC  

That means one gram of water require heat 4.18 j  to raise the temperature of water by one degree centigrade.

 Q-2  Determine specific heat in J/g times degree celcius . Mass of substance 29.2g, initial temp 34.4degrees celcius, final temp 68.1degrees celcius, heat applied 436.9 j, what is the formula of hear capacity ?

 Answer

Formula for the heat Q = M*C*delta T

                  M = mass of the substance in gram

                  C = specific hear capacity of substance

          delta T = final temperature  - initial temperature

from the problem

           delta T = 68.1 - 34.4

                      =33.7 C

         Mass M = 29.2 gram

         Heat Q  = 436.9j

plug the value in above formula we get

                    Q = M*C*delta T

            436.9 j = 29.2 gram * C* 33.7degree Celsius

              436.9 = 984.04gram degree celcius *  C

divide by 984.04gram degree Celsius we get

                 

 0.44 j / gram degree celcius = C (specific heat capacity )

 Answer heat capacity = 0.44 j/g K  or 0.44j/gC

heat capacity of water

density of table salt is 2.16g/cm^3. What is volume in Liters of a 154.3kg salt sample

density of table salt

Answer :-

   step - 1

             convert mass in grams 

                  1 kg = 1000  gram

          so 

            154.3kg  = 154000 gram 

   Step - 2 

             volume = mass /density 

                         =  154300 gram / 2.16 (g/cm^3) 

                         = 71435 cm^3 

   Step - 3

        1000 cm^3 = 1 liter 

       so

     71435 cm^3  = 71435/1000 liter 

                          = 71.44 liter

          Answer  = 71.44 liter

Balance the ionic equation by half reaction method O3+ Br- -------> O2 + BrO-

Answer:-

Step -1  

       Find the oxidation number of Br in BrO-

       in combination state oxygen has oxidation number = - 2 

             Br + O = - 1 

               Br -2 = -1 

                   Br = +1  

Step -2 

        write oxidation half reaction 

               Br-  <-----------> BrO- 

               -1                       +1

       here oxidation number of Br is increasing by 2 

       so add 2 electron to balance it 

       we get 

              Br-  <-----------> BrO-   + 2e-             

Step -3 

add OH - ions to balance the charge reaction 

Br- +  2OH-   <----------->BrO-   + 2e

Step - 4 

add H2O to balance the hydrogen and oxygen 

Br- +  2OH-   <----------->BrO-   +H2O +  2e 

balanced oxidation half reaction

 Br- +  2OH-   <----------->BrO-   +H2O +  2e   ---------------(1)

similarly you can get balanced half reduction reaction 

Balanced reaction half reaction 

O3 + H2O + 2 e- <-----------> O2 + 2 OH-        ---------------(2)

add both equations we get 

Br- +  2OH-  + O3 + H2O + 2 e- <-----------> BrO-   + 2e + O2 + 2 OH-

simplify the equation we get

O3+  Br- <----------->O2 +  BrO-

Answer : -  Equation was already balanced 

Balance the ionic equation

how many grams are in 1.8 x 10^23 atoms of silver

number of moles of silver = number of atoms / Avogadro number 

                                       = 1.8x10^23 atom/ 6.022x10^23 per moles

                                       =0.2989 moles 

mass in grams of silver    = number of moles * molar mass of substance 

                                       = 0.2989 moles * 107.87 gram /moles 

                                       = 32.34 grams 

Answer = 32.34 grams 

how many grams are in 1.8 x 10^23 atoms of silver

What is the molarity of pure water at 20 C?density table of water

 molarity of pure water

   density of water is at 20 C = 0.9982071 gram /ml

let we take 1 liter water 

                                   1 liter = 1000 ml 

            mass of 1 liter water = volume *density  

                                             = 1000 ml * 0.9982071 gram /ml 

                                             =  998.2071 gram

  molar mass of water  H2O = 2* 1.008 + 15.999

                                             = 18.015 gram / moles

number of moles of water in one liter 

                                            = mass of substance / molar mass

                                            = 998.2071 gram/18.015 (gram / moles)

                                            =55.41 moles 

        molarity of water        = number of moles of water / volume of water 

                                           = 55.41 moles/ 1liter 

                                           = 55.41moles/liter

Answer:-  molarity of water at 20 C is 55.41M

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?

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?

Answer:- 

                added volume = 0.05 /1000  

                                        = 0.00005 liter

    number of add moles  = molarity * volume 

                                        = 6moles / liter  * 0.00005 liter 

                                        =0.0003 moles 

total volume of solution   = 1.0 liter  +  0.00005 liter 

                                         = 1.00005 liter 

molarity of new solution = number of moles / volume in liter 

                                        =0.0003/1.00005 

                                        = 2.99x10^-4

    pH is this new solution= - log [H+]

Reaction is HCl  ----------> H+   +   Cl -

                            so [H+] = 2.99x10^-4

                         hence PH =  - log 2.99x10^-4

                                          =  3.52 

Hence there are big change in pH value

Answer  :- yes there are significant change in pH value

what is the approximate expected pH of a 0.030 M HNO3 solution?

Answer:- 

 HNO3, HCl ,H2SO4 are the  strong acid .

The formula for strong acid is  

                                                 pH = - log [H+] 

    

Dissociation reaction of HNO3 is

                                          HNO3 ---> H+    + NO3-

             So concentration of [H+]  = 0.030M 

                                             pH = - log 0.030 

                                            pH = 1.52

Answer pH = 1.52

what is the approximate expected pH of a 0.030 M HNO3 solution?

 

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?

Answer :-

number of moles of NaOH = molarity * volume in liter 

                                           =0.1298 moles / liter * (12.23/1000) liter  

divide by 1000 to convert ml to liter 

                                          =  0.001587454 moles

same of monoprotic acid so it will give one H + 

and base NaOH  gives only one OH - ions 

and net reaction will  H+   +    OH - ----> H2O 
 

 so moles of acid  = moles of base 

                             = 0.001587454 moles 

          molar mass = mass of substance / number of moles 

                             = 0.2602g/0.001587454 moles 

                             = 163.91 gram / moles 

 Answer :- molar mass of acid is 163.91 grams /mole

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?

10.00 mL of vinegar (mass=10.05 g) requires 14.77 mL of 0.4926 M NaOH to reach the end point. Calculate the molarity and mass percent of the acetic acid in the vinegar.

 Solution:-

Calculation of molarity :-

                  at the end point base  M1*V1*N1 = M2*V2*N2

          molarity of vinegar acid (M1) = to find

                      volume of vinegar V1 = 10 ml 

       It gives one H + ion  so that N1= 1 

                   Molarity of NaOH M2 = 0.4926

                           volume of NaOH = 14.77 ml 

                 It gives one OH- so N2 = 1   

                                                                                    Now use the formula  

                                 M1*V1*N1 = M2*V2*N2 

                                                                                   plug the values in this formula we get 

                                    M1*10*1 = 0.4926*14.77*1 

                                         10 M1 = 7.275702

                                              M1 = 0.7276 M

                                                              Answer molarity of vinegar is 0.7276 M 

Calculation of mass % :-

number of moles of base  = Molarity * volume in liter  of base 

                                       =(0.4926 moles /liter)*(14.77ml /1000)    

                                                                                        divided by 1000 to convert volume ml to  liter

                                       = 0.007275702 moles

reaction between acid and base 

                         CH3COOH + NaOH ---> CH3COONa  + H2O 

                                                                  here one moles of acid is reacting with 1 moles of base 

so number of moles of acid = number of moles of base 

                                         =  0.007275702 moles 

 Molar mass of acetic acid  CH3COOH = 60.05 g / moles

Mass of vinegar is = molar mass * number of moles 

                            =60.05 g/moles  * 0.007275702 moles 

                           =  0.4369059051 gram 

grams % of vinegar =  mass of acetic acid * 100/ total mass vinegar 

                              = 0.4369059051*100/ 10.05 % 

                              = 4.35%

 Answer mass % of vinegar is 4.35%

10.00 mL of vinegar (mass=10.05 g) requires 14.77 mL of 0.4926 M NaOH to reach the end point. Calculate the molarity and mass percent of the acetic acid in the vinegar.

Q-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?

solution :-

The titration of 10.00 mL of a diprotic acid solution

we get volume of given diprotic acid (V1) = 10 ml 

diprotic acid always gives 2 H + ions 

                          so number of of H + (N1) = 2

                   molarity of diprotic acid (M1) = ?

21.37 mL of a 0.1432 M NaOH solution

NaOH gives only one OH ions 

                                           so value of N2 = 1 

                                                 Volume V2 = 21.37 ml

                                        and molarity M2 = 0.1432 M

Formula for the reaction   

                                               M1*V1* N1= M2*V2*N2

          plug all values we get 

                                        M1 * 10.00ml*2 = 0.1432 *21.37 ml * 1 

          solve this condition we get 

                                                             M1 = 0.1530 M 

Answer :-   molarity of diprotic acid is 0.1530 M 

Q-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?

 

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