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A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added, and the

BARSIC [14]

Answer:

Rb+

Explanation:

Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.

Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.

Thus our calculations are:

V = 17.0 mL x 1 L / 1000 mL = 0.017 L

2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol

0.0425 mol = 4.36 g/ MW XOH

MW of XOH = (atomic weight of X + 16 + 1)

so solving the above equation we get:

0.0425 = 4.36 / (X + 17 )

0.7225 +0.0425X = 4.36

0.0425X = 4.36 -0.7225 = 3.6375

X = 3.6375/0.0425 = 85.59

The unknown alkali is Rb which has an atomic weight of 85.47 g/mol

6 0

3 years ago

For each row in the table below, decide whether the pair of elements will form a molecular or ionic compound. If they will, then

Olin [163]

Answer:

$\begin{array}{cccll}\textbf{Element 1} & \textbf{ Element 2} &\textbf{Compound?} &\textbf{Formula} &\textbf{Type}\\\text{Ar}&\text{Xe} &\text{No} &\text{None}&\text{Neither}\\\text{F}& \text{Cs} &\text{Yes} &\text{CsF} &\text{Ionic}\\\text{N} &\text{Br} &\text{Yes} & \text{NBr}_{3}&\text{molecular} \\\end{array}$