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:
Bohr's atomic model may have not been the accurate atomic model we have in the present, but he helped paved the way for accurate discoveries. His model is also called the planetary model. The nucleus, containing the neutrons and protons are situated at the center of the atom. The electrons are orbiting around the nucleus. The model is illustrated as shown in the attached picture.
Answer:
First one is 5.0 M ammonia and the Second one ?
Explanation:
Answer:
44Kj
Explanation:
These are the equations for the reaction described in the question,
Vaporization which can be defined as transition of substance from liquid phase to vapor
H2(g)+ 1/2 O2(g) ------>H2O(g). Δ H
-241.8kj -------eqn(1)
H2(g)+ 1/2 O2(g) ------>H2O(l).
Δ H =285.8kj ---------eqn(2)
But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as
H2O(l) ------>>H2O(g)---------------eqn(3)
But the equation from eqn(2) the eqn does go with vaporization so we can re- write as
H2O ------> H2(g)+ 1/2 O2(g)
Δ H= 285.8kj ---------------eqn(4)
To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)
Δ H=285.8kj +(-241.8kj)= 44kj
