They are considered the basic units of matter because all mass is made of atoms.
<span>7.379 * 10^(-4) is measured, hence prone to error, either human error or via measuring device. In this case,
100 cm = 1 m is written in stone and is unquestionable.
The density of the gold is 19.3 g/cm^3 and could be an approximation.
The approximation is good to at least one night.</span>
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
1) number of moles of N2 = n/2
2) Number of moles of CH4 = n/2
3) Total number of moles of the mixture = n/2 + n/2 = n
4) Kg of N2
mass in grams = number of moles * molar mass
molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol
=> mass of N2 in grams = (n/2) * 28 = 14n
mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg
Answer: mass of N2 in kg = 0.014n kg
Answer:- oxygen.
Explanations:- The electronic configuration is given and we are asked to figure out the electrically neutral atom that will have the electron configuration, 
.
The sum of electrons for this electron configuration is 8. If we look at the periodic table then 8 is the atomic number of oxygen.
So, the electrically neutral atom for the given electron configuration is oxygen.
