Answer:
sodium fusion is changed to sodium sulphide if sulphur is present in the compound
QUICK ANSWER
J.J. Thomson's cathode ray experiment was a set of three experiments that assisted in discovering electrons. He did this using a cathode ray tube or CRT. It is a vacuum sealed tube with a cathode and anode on one side.
The lattice energy of the compounds is distributed in the following decreasing order of magnitude: MgO > CaO > NaF > KCl.
<h3>KCl or NaF, which has a higher lattice energy?</h3>
The lattice energy increases with increasing charge and decreasing ion size.(Refer to Coulomb's Law.)MgF2 > MgO.Following that, we can examine NaF and KCl (both of which have 1+ and 1-charges), as well as atomic radii.NaF will have a larger LE than KCl since Na is smaller then K and F was smaller than Cl.
<h3>MgO or CaO, which has a larger lattice energy?</h3>
MGO is more difficult than CaO, hence.This is because "Mg" (two-plus) ions are smaller than "Ca" (two-plus) ions in size.MgO has higher lattice energy as a result.
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Answer: 8556 mm, or 855.6 cm (8560 mm to 3 sig figs)
Explanation: Convert mm to cm by dividing by 10 (1cm/10mm)
Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2
Calculate the volume occupied by 1.40 kg of foil in cm^3. 1.40kg = 1400g
1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au
Volume = Area (of the face) * Length
We want Length:
Length = Volume/Area
L = (518.5 cm^3/0.606 cm^2)
L = 855.6 cm (8556 mm) Round to 3 sig figs (856 cm and 8560 mm)
1a. calcium chloride (CaCl2)
b. 2HCl (aq) + Ca(OH)2 (s) —> CaCl2 (aq) + 2H2O (l)
i’m not sure about the rest but i hope this helped ^^
