A 12.0 g sample of HF is dissolved in water to give 3.1 \times 10^23.1×10 2mL of solution. The concentration of the solution is:

Explanation:

We can start with the data given by the problem:

Mass of HF = 12.0 g

Volume of solution = mL

If we want to calculate the "concentration", we have to remember the molarity equation:

So, we need to know the moles and the litters if we want to calculate the molarity.

Calculation of Litters

We already have a volume value, but the value given by the problem has "mL". So, we have to do a conversion. In 1 L we have 1000 mL (1 L = 1000 mL), therefore:

Calculation of moles

In this case, the compound is HF. So, we need to know the molar mass of HF if we want to do the conversion from grams to moles. The atomic mass of "F" is 18.99 g/mol and the atomic mass of "H" is 1 g/mol. So:

18.99 + 1 = 20 g /mol

In other words, 1 mol of HF is equal to 20 g (1 mol = 20 g HF). With this in mind, we can calculate the moles in the 12 g:

Calculation of the concentration of the solution

With these values know we can calculate the molarity:

I hope it helps!

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

put the answers   and i could answer.