Write the symbol for every chemical element that has atomic number greater than 9 and otomic mass less than 23.4 U

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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

Radon-222 or rn-222 has a half-life of 3.82 days. to determine the fraction of an original sample of rn-222 that remains unchanged after 7.646 days, we calculate the amount of rn-222 remaining after 7.646 days and divide it with the original amount. using 1 gram as a basis and using half-life calculation methods, we can determine that  0.249 or 0.25  grams of rn-222 will remain, and thus the fraction remaining is 0.25 or 1/4.