Answer:
The elements in Group 1 (lithium, sodium, potassium, rubidium, cesium, and francium) are called the alkali metals. All of the alkali metals have a single s electron in their outermost principal energy. ... For example, the electron configuration of lithium (Li), the alkali metal of Period 2, is 1s22s1.
The number of grams of Cl2 formed when 0.115 moles of HCl react with O2 is calculated as follows
write the chemical equation
4HCl + O2 -----> 2H2O + 2Cl2
if 0.115 moles of HCl reacted and from the equation above the reacting ratio between HCl and Cl2 is 4:2 therefore the moles of Cl2= 0.115 x2/4=0.0575moles
mass of Cl2= moles x molar mass of Cl2(35.5 x2=71g/mol)
=0.0575 moles x71g/mol = 4.0825 grams
I have provided a pair of spectra, NMR and IR for one of the compounds in the question.
The NMR spectrum provides the least information, as there is only a single peak that is a singlet. All three compounds have a methyl group adjacent to a carbonyl that would appear as a singlet. However, the chemical shift is closer to 3 ppm which suggests it is being deshielded further compared to a standard carbonyl which would appear at 2 ppm. This suggests the structure is most likely 1,1,1-trichloropropanone.
Looking at the IR spectrum, we see the carbonyl stretch at approximately 1750 cm⁻¹. We also see the presence of strong stretches at 750 and 850 cm⁻¹ which are very characteristic of a C-Cl stretch. Therefore, there is enough evidence to suggest that these spectra correspond to 1,1,1-trichloropropanone.
C. Temperature of the chamber. As temperature increases, pressure increases. As temp decreases, pressure decreases.