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Novosadov [1.4K]
3 years ago
11

Consider the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) Using standard thermodynamic data at 298K, calculate the entropy chang

e for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions. S°surroundings = J/K An error has been detected in your answer. Check
Chemistry
1 answer:
balandron [24]3 years ago
5 0

Answer:

the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.

Explanation:

3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)

∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K

∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.

For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,

then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.

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Im hvaing a hard time getting the right answer
Tanya [424]

Answer:

V=23.9mL

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

n=2.49g*\frac{1mol}{125.55 g}=0.0198mol

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

M=\frac{n}{V}\\\\V=\frac{n}{M}

By plugging in the moles and molarity, we obtain:

V=\frac{0.0198mol}{0.830mol/L}=0.0239L

Which in mL is:

V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL

Best regards!

6 0
3 years ago
PLEASE HELP...............................................
Serjik [45]
Umm its not a i think... lemme see lol

5 0
3 years ago
What is the concentration of each type of monosaccharide in lactose
polet [3.4K]

Answer:

Lactose comprises the monosaccharides glucose and galactose and maltose, comprising two glucose molecules, which occurs in barley, wheat, and malt.

Explanation:

5 0
2 years ago
The volume of a pond being studied for the effects of acid rain is 35 kiloliters (kL). There are 1,000 liters (L) in 1 kL and 1
Lerok [7]

Answer:

35,000,000,000 mL

Explanation:

You first multiply 35 times 1000.

35,000 L

Now you multiply 35,000 times 10^6

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4 0
3 years ago
The table shows the amount of radioactive element remaining in a sample over a period of time.
Assoli18 [71]

It would take 147 hours for 320 g of the sample to decay to 2.5 grams from the information provided.

Radioactivity refers to the decay of a nucleus leading to the spontaneous emission of radiation. The half life of a radioactive nucleus refers to the time required for the nucleus to decay to half of its initial amount.

Looking at the table, we can see that the initial mass of radioactive material present is 186 grams, within 21 hours, the radioactive substance decayed to half of its initial mass (93 g). Hence, the half life is 21 hours.

Using the formula;

k = 0.693/t1/2

k = 0.693/21 hours = 0.033 hr-1

Using;

N=Noe^-kt

N = mass of radioactive sample at time t

No = mass of radioactive sample initially present

k = decay constant

t = time taken

Substituting values;

2.5/320= e^- 0.033 t

0.0078 = e^- 0.033 t

ln (0.0078) = 0.033 t

t = ln (0.0078)/-0.033

t = 147 hours

Learn more: brainly.com/question/6111443

7 0
2 years ago
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