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Reil [10]
3 years ago
14

Write a balanced equation for the reaction of Al(H2O)63+ in aqueous KF. Include the physical states of each reactant and product

in your equation. Use an equilibrium arrow between reactants and products. Ignore potassium ions.
Chemistry
1 answer:
olga2289 [7]3 years ago
3 0

Explanation:

The potassium fluoride will dissociate into potassium ions and fluoride ions in their aqueous solution.

KF(aq)\rightarrow K^+(aq)+F^-(aq)

So, when 1 mol of  hexaaqua aluminium (III) reacts with 6 moles of fluoride ion it gives 1 mole of hexafluoroaluminate(III).

The reaction is given as:

Al(H_2O)_6^{3+}(aq)+6F^-(aq)\rightleftharpoons AlF_6^{3-}(aq)+6H_2O(l)

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Answer:

c

Explanation:

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Which is a characteristic of nuclear fusion but NOT nuclear fission?
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The answer is B. is the energy source of stars.

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BRAINLIEST ANSWER _ A balloon occupies a volume of 0.93 liters at a pressure of 84.5 kPa. When the balloon is allowed to expand
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3 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed.Drag the appropriate items to their respective b
Sonja [21]

Answer:

Ag+(aq)+Br−(aq)→AgBr(s)                                NEGATIVE

CaCO3(s)→CaO(s)+CO2(g)2                           POSITIVE

NH3(g)→N2(g)+3H2(g)                                    POSITIVE

2Na(s)+Cl2(g)→2NaCl(s)                                 NEGATIVE

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)           POSITIVE

I2(s)→I2(g)                                                        POSITIVE

Explanation:

We have to remember, to solve this problem, that the entropy of a gas is higher than that of a liquid which in turn  is higher than the solid. Therefore, comparing the reactants and products look for changes in the state of reactants and products. We also have to look for the increase or decrease of moles of each state based on the balanced chemical reaction.

Ag+(aq)+Br−(aq)→AgBr(s)

The reaction product is a single solid and the  the reactants were 2 species in solution. The change in entropy is negative.

CaCO3(s)→CaO(s)+CO2(g)2

Here we have a solid reactant and we have a solid product plus a gas product. The change in entropy is positive.

NH3(g)→N2(g)+3H2(g)

We have 4 mole gases as products starting from 1 mol reactant gas, the entropy has increased.

2Na(s)+Cl2(g)→2NaCl(s)

In this reaction 2 mol solid Na and 1 mol Cl₂ gas are converted into 2 mol solid NaCl, the entropy has decreased.

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)

The products are 7 mol of gas versus 6 mol of gas reactants and therefore entropy has increased.

I2(s)→I2(g)

1 mol solid I₂ goes into 1 mol gas making the change in  the entropy higher.

4 0
3 years ago
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