Chemistry, 15.04.2020 00:47 camdynmmm

In a coffee‑cup calorimeter, 65.0 mL of 0.890 M H 2 SO 4 was added to 65.0 mL of 0.260 M NaOH . The reaction caused the temperature of the solution to rise from 23.78 ∘ C to 25.55 ∘ C. If the solution has the same density as water (1.00 g/mL) and specific heat as water (4.184 J/g‑K), what is Δ H for this reaction (per mole of H 2 O produced)? Assume that the total volume is the sum of the individual volumes.

Answers: 2

Show answers

Answers

Answer from: zaire8

Answer : The enthalpy of neutralization is, 56.96 kJ/mole

Explanation :

First we have to calculate the moles of H₂SO₄ and NaOH.

$\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.890mole/L\times 0.065L=0.0578mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.260mole/L\times 0.065L=0.0169mole$

The balanced chemical reaction will be,

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

From the balanced reaction we conclude that,

As, 2 mole of NaOH neutralizes by 1 mole of H₂SO₄

So, 0.0169 mole of NaOH neutralizes by 0.00845 mole of H₂SO₄

That means, NaOH is a limiting reagent and H₂SO₄ is an excess reagent.

Now we have to calculate the moles of H₂O.

As, 2 mole of NaOH react to give 2 mole of H₂O

So, 0.0169 mole of NaOH react to give 0.0169 mole of H₂O

Now we have to calculate the mass of water.

As we know that the density of water is 1.00 g/ml. So, the mass of water will be:

The volume of water = 65.0ml+65.0ml=130ml

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/ml\times 130ml=130g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

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

m = mass of water = 130 g

$T_{final}$ = final temperature of water = 23.78^oC

$T_{initial}$ = initial temperature of metal = 25.55^oC

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

$q=130g\times 4.184J/g^oC\times (25.55-23.78)^oC$

q=962.7J

Thus, the heat released during the neutralization = -962.7 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -962.7 J

n = number of moles used in neutralization = 0.0169 mole

$\Delta H=\frac{-962.7J}{0.0169mole}=-56964.49J/mole=-56.96kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.96 kJ/mole

Answer from: Quest

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.