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3 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]3 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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FromTheMoon [43]

Answer:

the wavelength, in nm, of the photon is 487.5 nm

Explanation:

Given:

n = 4 (excited)

n = 2 (relaxes)

Question: Calculate the wavelength, in nm, λ = ?

First, it is important to calculate the energy of the electron when it excited and then when it relaxes.

$E_{1} =\frac{-13.6}{n^{2} } =\frac{-13.6}{4^{2} } =-0.85eV$ (excited)

$E_{2} =\frac{-13.6}{2^{2} } =-3.4eV$ (relaxes)

The change of energy

ΔE = E₁ - E₂=-0.85 - (-3.4) = 2.55 eV = 4.08x10⁻¹⁹J

For a photon, the wavelength emitted

$\lambda =\frac{hc}{delta(E)}$

Here

h = Planck's constant = 6.63x10⁻³⁴J s

c = speed of light = 3x10⁸m/s

Substituting values:

$\lambda =\frac{6.63x10^{-34}*3x10^{8} }{4.08x10^{-19} } =4.875x10^{-7} m=487.5nm$

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3 years ago

I need help with acids and bases.

Alisiya [41]

levels greater than seven are bases

levels on 7 are neutral

levels low than 7 are acids

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3 years ago

How many grams of Cl2 must react to produce 0.0923 mol of AlCl3? Show work.

alexandr402 [8]

First, let's state the chemical reaction:

$2Al+3Cl_2\to2AlCl_3\text{.}$

We can find the number of moles of Cl2 required to produce 0.0923 moles of AlCl3, doing a rule of three: 3 moles of Cl2 reacted produces 2 moles of AlCl3:

$\begin{gathered} 3molesCl_2\to2molesAlCl_3 \\ \text{?moles Cl}_2\to0.0923\text{ moles }AlCl_3\text{.} \end{gathered}$

The calculation would be:

$0.0923molesAlCl_3\cdot\frac{3molesCl_2}{2molesAlCl_3}=0.138molesCl_2.$

And the final step is to convert this number of moles to grams. Remember that the molar mass can be calculated using the periodic table, so the molar mass of Cl2 is 70.8 g/mol, and the conversion is:

$0.138molesCl_2\cdot\frac{70.8gCl_2}{1molCl_2}=9.770gCl_2.$

The answer is that we need 9.770 grams of Cl2 to produce 0.0923 moles of AlCl3.

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1 year ago

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3 years ago

What is the final product of the following sequence of reactions? A. A B. B C. C D. D 21. Which of the following reacts the fast

Gala2k [10]

Answer:

Correct answer is Option A:

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4 years ago