Answer:
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
Explanation:
...[1]
..[2]
..[3]
The standard enthalpy of formation of ethanoic acid :
..[4]
Using Hess's law to calculate :
2 × [1] + 2 × [2] - [3] = [4]
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
The volume of sample = 2.25(1.5)(2.25) = 7.59375 cm3
Density = mass / volume
Density = 55.75 / 7.59375 = 7.34156 g/cm3
As the density of iron is 7.86 g/cm3 instead, it is not iron
Answer:
B. burning a piece of wood
Explanation:
The Chemical Would Be The Air Coming From The Wood While Burning It
Tell Me If Im Correct
Answer:
35.8 u
Explanation:
The atomic mass of Cl is the weighted average of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its percent abundance).
Atomic mass of Cl-35 = 17p + 18n = 17 × 1.007 u + 18 × 1.009 u
= 17.119 u + 18.162 u = 35.28 u
Atomic mass of Cl-37 = 17p + 20n = 17 × 1.007 u + 20 × 1.009 u
= 17.119 u + 20.180 u = 37.30 u
Set up a table for easy calculation.
0.755 × 35.28 u = 26.64 u
0.245 × 37.30 u = 9.138 u
TOTAL = 35.8 u
Note: The actual atomic mass of Cl is 35.45 u.
The calculated value above is incorrect because
(a) the given isotopic percentages are incorrect and
(b) the protons and neutrons have less mass when they are in the nucleus than when they are free. Thus, the calculated masses of Cl-35 and Cl-37 are too high.
