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Bond [772]
3 years ago
13

The decomposition of 57.0 g of Fe2O3 results in Consider the following reaction. 2Fe2O3 ---> 4Fe + 3O2 deltaH degree rxn = +

824.2 kJ decomposition of 57.0 g of Fe2O3 results in the release of 294 kJ of heat. A. the absorption of 23500 kJ of heat. B. the absorption of 147 kJ of heat. C. the absorption of 294 kJ of heat. D. the release of 23500 kJ of heat. E. the release of 147 kJ of heat.
Chemistry
1 answer:
lozanna [386]3 years ago
3 0

Answer:

The correct answer is option E.

Explanation:

2Fe_2O_3\rightarrow 4Fe + 3O_2 \Delta H^o_{rxn} = 824.2 kJ

Mass of ferric oxide decomposed = 57.0 g

Moles of ferric oxide decomposed = \frac{57.0 g}{160 g/mol}=0.35625 mol

According to reaction, 2 moles ferric oxide on combustion gives 824.2 kJ of heat.

Then heat given by on decomposition of 0.35625 mol of ferric oxide will be:

\frac{824.2 kJ}{2}\times 0.35625 =146.81 kJ\approx 147 kJ

So, on decomposition of 57.0 grams of ferric oxide 147 kilo Joules of heat was released.

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B would be correct if it had another carbon.

A is like C but incorrect for the same reason. 1,2-dibromoethane is what A is.

The answer has to be D where the bromines are right across from each other.

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3 years ago
What is the speed 0.36<br> km/h in cm/s
azamat

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It's 10 cenimeters per second

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Nitrogen monoxide, NO, is formed in automobile exhaust by the reaction of nitrogen and oxygen in the air: N2 (g) + O2 (g) ↔ NO (
Solnce55 [7]

Answer:

The equilibrium concentration of NO is 0.001335 M

Explanation:

Step 1: Data given

The equilibrium constant Kc is 0.0025 at 2127 °C

An equilibrium mixture contains 0.023M N2 and 0.031 M O2,

Step 2: The balanced equation

N2(g) + O2(g) ↔ 2NO(g)

Step 3:  Concentration at the equilibrium

[N2] = 0.023 M

[O2] = 0.031 M

Kc = 0.0025 = [NO]² / [N2][O2]

Kc = 0.0025 = [NO]² / (0.023)(0.031)

[NO] = 0.001335 M

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6 0
3 years ago
25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?
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Answer:

pH = 11.95≈12

Explanation:

Remember  the reaction among aqueous acetic acid (CH_3COOH) and aqueous sodium hydroxide (NaOH)

CH_3COOH + NaOH ->  CH_3COONa + H_2O

First step. Need to know how much moles of the substances are present

0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH = 0.0025 mol NaOH

0.003 mol NaOH * 1 mol CH_3COOH/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

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[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95

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