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2 years ago

2) What is the change in enthalpy when 3.400 moles of aluminum react with excess ammonium nitrate (NH4NO3)

2 answers:

5 0

The enthalpy change with the reaction of 3.4 moles of aluminum in the reaction is -3045 kJ. The reaction is exothermic in nature.

<h3>What is an exothermic and endothermic reaction?</h3>

The reaction in which the energy is released and the value of change in enthalpy is negative is an exothermic reaction. The reaction in which the energy is absorbed with the positive value of change in enthalpy.

For a balanced chemical reaction, for 2 moles of Aluminum reaction, the change in enthalpy has been -2030 kJ. The change in enthalpy for the reaction of 3.4 moles of aluminum is given as:

$\rm 2\;mol\;Al=-2030 \;kJ\\\\3.4\;mol\;Al=\dfrac{-2030}{2}\;\times\;3\;kJ\\\\ 3.4\;mol\;Al=-3045\;kJ$

The change in enthalpy for 3.4 moles of aluminum reaction is -3045 kJ. The reaction is exothermic in nature.

Learn more about exothermic reaction, here:

brainly.com/question/10373907

Vlad [161]2 years ago

3 0

Answer: See below

Explanation:

(i) 2AI + 3 NHANO3 → 3 N2+ 6 H2O + Al2O3 ΔH = -2030kJ

From stoichiometric coefiicients, it can be seen that:

When 2 moles of Al react, change in enthalpy = -2030kJ

So, when 3.4 moles of Al react, change in enthalpy = -(2030/2) x 3.4 kJ = -3451 kJ

Therefore, the change in enthalpy is -3451 kJ

(ii) The change in enthalpy is negative. This means heat is released during the reaction.

So the reaction is exothermic

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3 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

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3 years ago

A sample of sugar contains 1.505 × 1023 molecules of sugar. How many moles of sugar are present in the sample?

alisha [4.7K]

1 mole ------------ 6.02 x10²³ molecules
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3 years ago