Conversion of mole to grams
k in mole = 1 mole/ atomic mass
K in mole =1/ 39.0983 g/mole
= 0.255765 g/mole
converting 40 grams of K
K 40 grams x [ 1 mole/ 39.0983 grams] = 1.0230623 mole
There are 1.0230623 moles of K in 40 K of Potassium
Answer:
The empirical formula of this substance is:

Explanation:
To find the empirical formula of this substance we need the molecular weight of the elements Carbon, Hydrogen and Oxygen, we can find this information in the periodic table:
- C: 12.01 g/mol
- H: 1.00 g/mol
- O: 15.99 g/mol
With the information in this exercise we can suppose in 100 g of the substance we have:
C: 48.64 g
H: 8.16 g
O: 43.2 g (100 g - 48.64g - 8.16g= 43.2 g)
Now, we need to divide these grams by the molecular weight:

We need to divide these results by the minor result, in this case O=2.70 mol

We need to find integer numbers to find the empirical formula, for this reason we multiply by 2:

This numbers are very close to integer numbers, so we can find the empirical formula as subscripts in the chemical formula:

A) James Cook.
B) He put his sailors on a strict diet to see if they would get scurvy.
C) Sauerkraut.
D) He told others of this diet and that none of his sailors died of scurvy.
E) Chemicals can be found almost anywhere and almost anyone can be a scientist.
Answer : The vapor pressure (in atm) of a solution is, 0.679 atm
Explanation : Given,
Mass of
= 1.00 kg = 1000 g
Moles of
= 3.68 mole
Molar mass of
= 18 g/mole
Vapor pressure of water = 0.692 atm
First we have to calculate the moles of
.

Now we have to calculate the mole fraction of 

Now we have to partial pressure of solution.
According to the Raoult's law,

where,
= vapor pressure of solution
= vapor pressure of water = 0.692 atm
= mole fraction of water = 0.938



Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm
Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:


Now to calculate the molarity of original solution:


Thus the molar concentration of sulfuric acid in the original sample is 1.943 M