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3 years ago

What mass of xenon tetrafluoride, xef 4 , has the same number of fluorine atoms as 25.0 g of oxygen difluoride, of2?

Chemistry

2 answers:

Pavlova-9 [17]3 years ago

7 0

Answer is: mass of xenon tetrafluoride is 47.98 grams.

1) m(OF₂) = 25.0 g; mass of oxygen difluoride.

n(OF₂) = m(OF₂) ÷ M(OF₂).

n(OF₂) = 25 g ÷ 54 g/mol.

n(OF₂) = 0.462 mol; amount of substance.

In one oxygen difluoride (OF₂) there are two fluorine atoms: n(F) : n(OF₂) = 2 : 1.

n(F) = 0.462 mol · 2.

n(F) = 0.926 mol.

N(F) = n(F) · Na(Avogadro constant).

N(F) = 0.926 mol · 6.022·10²³ 1/mol.

N(F) = 5.576·10²³; number of fluorine atoms in oxygen difluoride.

2) In one molecule of xenon tetrafluoride there are four fluorine atoms:

n(XeF₄) : n(F) = 1 :4.

n(XeF₄) = 0.926 mol ÷ 4.

n(XeF₄) = 0.2315 mol; amount of xenon tetrafluoride.

m(XeF₄) = 0.2315 mol · 207.28 g/mol.

m(XeF₄) = 47.98 g.

TiliK225 [7]3 years ago

3 0

To determine the mass of xenon tetrafluoride, we need to know first the number of fluorine atoms present in <span>oxygen difluoride. We need to convert first the mass into moles then make use of the relation of the elements from the chemical formula. Then, use the avogadro's number to convert it to number of atoms. Then, we do the reverse of the steps above but this time for </span><span>xenon tetrafluoride.

25.0 g OF2 ( 1 mol / 54 g ) ( 2 mol F / 1 mol OF2 ) ( 6.022 x10^23 atoms F / 1 mol F ) ( 1 mol / 6.022x10^23 atoms) ( 1 mol XeF4 / 4 mol F ) (207.3 g / 1 mol XeF4) = 47.99 g XeF4</span>

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Is the solution containing 0.20 M Pb2 and 0.10 M Cl (Ksp 1.6x105) a) saturated, b) unsaturated, c) at equilibrium, d) none of th

raketka [301]

Answer : The correct option is, (b) unsaturated

Explanation :

Ionic product : It is defined as the product of the concentrations of the ions present in solution raised to the same power by its stoichiometric coefficient in a solution of a salt. This takes place at any concentration. The ionic product is represented as, Q.

Solubility product constant : It is defined as the product of the concentration of the ions present in a solution raised to the power by its stoichiometric coefficient in a solution of a salt. This takes place at equilibrium only. The solubility product constant is represented as, $K__{sp}$ .

Now we have to calculate the ionic product of given solution.

The balanced chemical reaction is,

$PbCl_2(aq)\rightarrow Pb^{2+}(aq)+2Cl^-(aq)$

The expression of ionic product will be,

$Q=[Pb^{2+}][Cl^-]^2$

$Q=(0.20)\times (0.10)^2$

$Q=0.002$

The ionic product of solution is, 0.002

There are three cases for the solubility :

When $Q$ this means that the solution is unsaturated solution and more solid will be dissolve.

When $Q=k_{sp}$ this means that the solution is saturated solution.

When $Q>k_{sp}$ this means that the solution is supersaturated solution and solid will be precipitate.

From this we conclude that the value of $k_{sp}=1.6\times 10^5$ is greater than the ionic product that means the solution is unsaturated solution and more solid will be dissolve.

Hence, the correct option is, (b) unsaturated

6 0

3 years ago

Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.

Softa [21]

Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

$PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

$K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

$[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

$[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

$K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains $[Br^{-}]_{o}$ = 0.10 and there will be no lead ions. So, $[Pb^{2+}]_{o}$ = 0

So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$ .