balance the oxidation reduction PbO2 yields Pb2+ + O2

Answer :-

oxidation number of oxygen in compound state is -2 

so the oxidation number of Pb in PbO2 

Pb + 2(-2)  = 0 

             Pb = 4 

and Pb +2 has oxidation number +2 

              change in oxidation number is 2-4 = - 2  (reduction )

oxidation number of oxygen in O2 is 0 

so change in oxidation of oxygen is  0- (2)  = 2 (oxidation )

and there are oxygen atom there

                so change in oxidation number is = 2* 2 

                                                                      = 4 

 there ratio of oxidation and reduction is  4  :  2   = 2  :   1  

cross multiply by these relation we get

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