Given chemical equations for these reactions s(s) + o2(g)  so2(g) ∆h˚ –296.8 kj•mol–1 h2(g) + ½ o2(g)  h2o(l) ∆h˚ –285.8 kj•mol–1 h2(g) + s(s)  h2s(g) ∆h˚ –20.6 kj•mol–1 what is the value of ∆h for the reaction below? 2 h2s(g) + 3 o2(g)  2 h2o(l) + 2 so2(g) (a) –603.2 kj•mol–1 (b) –562.0 kj•mol–1

Answers

Answer 1

Answer: Answer is: enthalpy is -1124 kJ/mol, if we divide reaction with two enthalpy is -562.0 kJ/mol.

Reaction 1: S(s) + O₂(g) → SO₂(g) ΔrH₁ = -296.8 kJ/mol.
Reaction 2: H₂(g) + ½ O₂(g) → H₂O(l) ΔrH₂ = -285.8 kJ/mol.
Reaction 3: H₂(g) + S(s) → H₂S(g) ∆rH₃ = -20.6 kJ/mol.
Reaction 4: 2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g) ΔrH₄ = ?
Using Hess's law reaction number 4 is sum of reaction number 1 multiply with two, reaction number 2 multiply with two and reaction 3 reversed and multiply with two:
ΔrH₄ = 2 · (-296.8 kJ/mol) + 2 · (-285.8 kJ/mol) + 2 · 20.6 kJ/mol.
ΔrH₄ = -1124 kJ/mol.

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Given: The equation for the synthesis of ammonia is H2(g) + N2(g) → NH3(g). If 12.0 grams of hydrogen are used up by the reaction, how many grams of ammonia will be produced?

Answers

Answer:

The answer to your question is 68 g

Explanation:

Data

mass of H₂ = 12 g

mass of NH₃ = ?

Balanced chemical reaction

3H₂ + N₂ ⇒ 2NH₃

Process

1.- Calculate the molar mass of H₂

H₂ = 1 x 6 = 6

NH₃ = 2[(14 x 1) + (3 x 1)]

= 2[14 + 3]

= 2[17]

= 34 g

2.- Use proportions to find the mass of NH₃

6 g of H₂ --------------- 34 g of NH₃

12g of H₂ ---------------- x

x = (12 x 34) / 6

x = 408/6

x = 68 g of NH₃

A sample of nitric acid contains both H3O ions and NO3 ions. This sample has a pHvalue of 1.
What is the color of methyl orange after it is added to this sample?

Answers

The indicator will lead to a solution that has a red color. Methyl orange is an indicator used in titrations because it changes colors at all pH values. When in acidic solution it is red and in basic it is yellow. A pH of 1 is acidic, therefore a red solution.

the empirical formula of a compound is NO2. it's molecular mass is 92g/mol. What is its molecular formula?

Answers

The molecular formula:
$N_x(O_2)_x$

$M_(N_x(O_2)_x)=x * M_N + x * 2 * M_O \n \nm_N=14 \ (g)/(mol) \nm_O=16 \ (g)/(mol) \n M_(N_x(O_2)_x)=92 \ (g)/(mol) \n \n92=x * 14+x * 2 * 16 \n92=14x+32x \n92=46x \n(92)/(46)=x \nx=2 \n \nN_2(O_2)_2 \Rightarrow N_2O_4$

The molecular formula is N₂O₄.

What is the molality of a solution

Answers

Answer:

The formula for molality is m = moles of solute / kilograms of solvent. In problem solving involving molality, we sometimes need to use additional formulas to get to the final answer. One formula we need to be aware of is the formula for density, which is d = m / v, where d is density, m is mass and v is volume

Explanation:

it is a measure of the concentration of a solute in a solution in terms of amount of substance in a specified amount of mass of the solvent. This contrasts with the definition of molarity which is based on a specified volume of solution.

The lone pair of electrons in ammonia allows the molecule to: A.assume a planar structure. B.act as an oxidizing agent. C.act as a Lewis acid in water. D.act as a Lewis base in water.

Answers

Answer:

The correct option is: D.act as a Lewis base in water.

Explanation:

Ammonia is a hydride of nitrogen with the chemical formula NH₃. It is a colorless gas with a characteristic pungent smell. According to the VSEPR theory, it has a trigonal pyramidal structure.

In water, ammonia acts as a Lewis base due to the presence of lone pair on the nitrogen atom. Lewis bases are electron pair or lone pair donors.

In Ernest Rutherford's gold foil experiments, some alpha particles were deflected from their original paths, but most passed through the foil with no deflection. Which statement about gold atoms is supported by these experimental observations?Gold atoms consist mostly of empty space.

Gold atoms are similar to alpha particles.

Alpha particles and gold nuclei have opposite charges.

Alpha particles are less dense than gold atoms.

Answers

According to ErnestRutherford's gold foil tests, the true statement is that space makes up the majority of gold atoms. The correct option is A.

According to Rutherford's studies, the majority of the alphaparticles encountered free space inside the atom since they mostly went through the gold foil with little to no deflection.

This finding helped scientists comprehend that atoms have a small, compact nucleus at their center and are primarily made up of empty space elsewhere in the atom.

The nuclearmodel of the atom was created as a result of the deflection of several alpha particles, which showed that they came into contact with the positively charged atom's nucleus.

Thus, the correct option is A.

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Your question seems incomplete, the probable complete question is:

A. Gold atoms consist mostly of empty space.

B. Gold atoms are similar to alpha particles.

C. Alpha particles and gold nuclei have opposite charges.

D. Alpha particles are less dense than gold atoms

Answer:

Explanation:

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