Which of the following pairs of elements are likely to form ionic compounds? check all that apply.
sodium and potassium
nitrogen and iodine
potassium and sulfur
chlorine and bromine
helium and oxygen
magnesium and chlorine

Answer : The correct pairs of elements likely to form ionic compounds are, potassium and sulfur, magnesium and chlorine.

Explanation :

Ionic compound : These are the compounds in which the atoms are bonded through the ionic bond. The ionic bond are formed by the complete transfer of the electrons.

That means, the atom which looses the electron is considered as an electropositive atom and the atom which gains the electron is considered as an electronegative atom.

And these bonds are formed between one metal and one non-metal.

From the given pairs of elements, potassium and sulfur, magnesium and chlorine are form ionic compound because potassium and magnesium are the metals and sulfur and chlorine are the non-metals. So, they can easily form ionic compound by the complete transfer of electrons.

While the other pairs, sodium and potassium are the metals, nitrogen and iodine, chlorine and bromine, helium and oxygen are the non-metals. They do not form ionic bond.

Hence, the correct pairs of elements likely to form ionic compounds are, potassium and sulfur, magnesium and chlorine.

Here is the link to the all information u need i mean answer key

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is  2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry  problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that     10 dm = 1 m  ⇔ 10³ dm³ = 1 m³then we use the fact that     10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting:     now we have to find the moles of hydrochloric  acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by    molar mass = (mass ca) + (mass c) + 3(mass o)    = 40 + 12 + 3(16)    = 100 g/molconvert the given 5 g to moles:     mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis    mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃)    = 0.05 mol hcl * 2    = 0.1 mol hcl  we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get     volume = moles / concentration    volume = 0.1 mol / (0.002 mol/cm³)    volume = 50 cm³