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Likurg_2 [28]
2 years ago
13

Calculate the amount of energy released in the formation of one mole of BaSe bonds (not lattice energy). The radius of the bariu

m ion is 1.35 Å, and the radius of the selenide ion is 1.98 Å. Note that 1Å=10−10m.
Chemistry
1 answer:
zysi [14]2 years ago
3 0
The correct answer for the question that is being presented above is this one: "2.02x10^3 kJ/mol." Calculate the amount of energy released in the formation of one mole of BaSe bonds (not lattice energy). The radius of the barium ion is 1.35 Å, and the <span>radius of the selenide ion is 1.98 Å. Note that 1Å=10−10m.</span>
You might be interested in
You place 36.5 ml of 0.266 M Ba(OH)2 in a coffee-cup calorimeter at 25.00°C and add 56.6 ml of 0.648 M HCl, also at 25.00°C. Aft
olya-2409 [2.1K]

Answer : The enthalpy of reaction (\Delta H_{rxn}) is, -96.9 kJ/mole

Explanation :

First we have to calculate the mass of solution.

Mass=Density\times Volume

Volume of solution = Volume of HCl + Volume of Ba(OH)_2

Volume of solution = 56.6 mL + 36.5 mL

Volume of solution = 93.1 mL

Density of solution = 1 g/mL

Mass=1g/mL\times 93.1mL=93.1g

The mass of solution is, 93.1 grams.

Now we have to calculate the heat released in the system.

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat released = ?

m = mass = 93.1 g

C_p = specific heat capacity of water = 4.184J/g^oC

T_1 = initial temperature  = 25.0^oC

T_2 = final temperature  = 29.83^oC

Now put all the given value in the above formula, we get:

Q=93.1g\times 4.184J/g^oC\times (29.83-25.00)^ioC

Q=1881.43J=1.88kJ        (1 kJ = 1000 J)

Now we have to calculate the moles of Ba(OH)_2 and HCl.

\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}

\text{Moles of }Ba(OH)_2=0.266M\times 0.0365L=9.71\times 10^{-3}mol

and,

\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}

\text{Moles of }HCl=0.648M\times 0.0566L=3.66\times 10^{-2}mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O

From the balanced reaction we conclude that

As, 1 mole of Ba(OH)_2 react with 2 mole of HCl

So, 9.71\times 10^{-3}  moles of Ba(OH)_2 react with 9.71\times 10^{-3}\times 2=0.0194 moles of HCl

From this we conclude that, HCl is an excess reagent because the given moles are greater than the required moles and Ba(OH)_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2O

From the reaction, we conclude that

As, 1 mole of Ba(OH)_2 react to give 2 mole of H_2O

So, 9.71\times 10^{-3}  moles of Ba(OH)_2 react with 9.71\times 10^{-3}\times 2=0.0194 moles of H_2O

Now we have to calculate the change in enthalpy of the reaction.

\Delta H_{rxn}=-\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 1.88 kJ

n = moles of reaction = 0.0194 mole

Now put all the given values in above expression, we get:

\Delta H_{rxn}=-\frac{1.88kJ}{0.0194mole}=-96.9kJ/mole

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, -96.9 kJ/mole

3 0
3 years ago
Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

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