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Soloha48 [4]
3 years ago
7

Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

(g)ΔH1=−241.8 kJ 2) X(s)+2Cl2(g)⟶XCl4(s)ΔH2=+356.9 kJ 2) X(s)+2Cl2(g)⟶XCl4(s)ΔH2=+356.9 kJ 3) 12H2(g)+12Cl2(g)⟶HCl(g)ΔH3=−92.3 kJ 3) 12H2(g)+12Cl2(g)⟶HCl(g)ΔH3=−92.3 kJ 4) X(s)+O2(g)⟶XO2(s)ΔH4=−639.1 kJ 4) X(s)+O2(g)⟶XO2(s)ΔH4=−639.1 kJ 5) H2O(g)⟶H2O(l)ΔH5=−44.0 kJ 5) H2O(g)⟶H2O(l)ΔH5=−44.0 kJ what is the enthalpy, ΔH,ΔH, for this reaction? XCl4(s)+2H2O(l)⟶XO2(s)+4HCl(g)
Chemistry
1 answer:
Bond [772]3 years ago
5 0

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

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Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
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The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

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f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

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<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

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Learn more here:brainly.com/question/15962928

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A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving 5.66 g of NH₄NO₃ and 4.42 g of (NH₄)₃ PO₄ in enoug
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In finding the molarity of a solution, we use the following formula:

M=moles soluteL solution

What is Molarity?

The number of moles of the solute is calculated by dividing the mass of the solute by its molar mass.

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The molar mass of  NH4NO3 and (NH4)3PO4 are  80.043 g/mol and 149.0867 g/mol, respectively.

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Pb has an atomic number of 82. It has an isotope Pb-207. The mass number is 207.
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Explanation:

Given data:

Atomic number of Pb = 82

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Number of protons = ?

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All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.

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