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