Which is a possible excited state configuration for Nitrogen?

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

A student reacted NiS2 (MM = 122.83 g/mol) with O2 (MM=32.00 g/mol) to make SO2 (MM=64.07 g/mol) according to this balanced equa

Ilya [14]

Answer:

Percentage yield = 0.49 × 10² %

Explanation:

Given data:

Actual yield of SO₂ = 4.309 ×10² g

Theoretical yield of SO₂ = 8.78 ×10² g

Percentage yield = ?

Solution:

Chemical equation:

2NiS₂ + 5O₂ → 2NiO + 4SO₂

Percentage yield:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield = 4.309 ×10² g / 8.78 ×10² g × 100

Percentage yield = 0.491 × 100

Percentage yield = 49.1%

In scientific notation:

Percentage yield = 0.49 × 10² %

5 0

3 years ago

The correct sequence where reactivity towards oxygen increases.

ziro4ka [17]

Answer:

Option D is good to go!

Explanation: as per the reactivity series more reactive substances will react with the counterpart substance.The most reactive substance here is calcium while the least reactive is aluminium, the magnesium comes in between.As per their reactivity, these substances will react with oxygen.

Explanation:

3 0

3 years ago

How many grams of sulfur must be burned to give 100.0 g of So2

andriy [413]

Answer:

50 g of S are needed

Explanation:

To star this, we begin from the reaction:

S(s) + O₂ (g) → SO₂ (g)

If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.

According to stoichiometry, we can determine the moles of sulfur dioxide produced.

100 g. 1mol / 64.06g = 1.56 moles

This 1.56 moles were orginated by the same amount of S, according to stoichiometry.

Let's convert the moles to mass

1.56 mol . 32.06g / mol = 50 g

4 0

2 years ago

Why does the fact that Americans are using less fertilizer not indicate a decline in American farming?

meriva

It shows that the population is moving towards a more organic and natural way of growing crops. Rather than using fertilizer and putting toxins on the plants that we eat from, we are getting smarter and growing food naturally. Thus not killing ourselves in the process.

4 0

3 years ago

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