What is the standard enthalpy of formation of liquid methylamine (CH3NH2) ?C(s)+O2(g) -> CO2(g); ?H=-393.5 kJ 2H2O(l) -> 2H2(g)+O2(g); ?H=571.6 kJ N2(g)+O2(g) -> NO2(g); ?H=33.10 kJ 4CH3NH2(l)+13O2(g) -> 4CO2(g)+4NO2(g)+10H2O(l); ?H=-4110.4 kJ The calculated answer is -47.3 kJ/mol. Show the work to confirm or deny the answer

ΔH =-47.3kJ

Explanation:

You must know standard enthalpy is defined as change of enthalpy during the formation of 1 mole of the substance from its constituent elements

You can find the standard enthalpy of any reaction from the sum of another similar reactions (Hess's law) as follows:

For methylamine, CH₃NH₂ is:

C(s) + 5/2 H₂(g) + 1/2 N₂(g) → CH₃NH₂ (l)

1. C(s) + O₂(g) → CO₂(g); ΔH=-393.5 kJ

2. 2H₂O(l) → 2H₂(g) + O₂(g); ΔH=571.6 kJ

3. 1/2N₂(g) + O₂(g) → NO₂(g); ΔH=33.10kJ

4. 4CH₃NH₂(l) + 13O₂(g) → 4CO₂(g) + 4NO₂(g) + 10H₂O(l); ΔH=-4110.4 kJ

The sum of (1)+(3) produce:

C(s) + 2O₂(g) + 1/2N₂(g) → CO₂(g) + NO₂(g) ΔH=-393.5kJ + 33.10kJ = -360.4kJ

-5/4 (2):

C(s) + 13/4O₂(g) + 1/2N₂(g) + 5/2H₂(g) → CO₂(g) + NO₂(g) + 5/2 H₂O(l)

ΔH= -360.4kJ -5/4 (571.6kJ) = -1074.9kJ

And this reaction -1/4 (4):

C(s) + 5/2 H₂(g) + 1/2 N₂(g) → CH₃NH₂(l)

ΔH= -1074.9kJ -1/4(-4110.4kJ)

Now, you can confirm the calculated answer!

is it a purebred? it looks more like a red nose, but it doesn't have the red could it be a staffordshire bull terrior? unsure if that's the right name for it sorry

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.