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Alekssandra [29.7K]
3 years ago
7

Explain why this reaction can be classified as a synthesis reaction

Chemistry
1 answer:
rusak2 [61]3 years ago
5 0
The reason why the reaction written on the picture can be classified as a synthesis reaction is :
the reaction shows one compound that formed from two compounds 

hope this helps
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Combining 0.350 mol Fe 2 O 3 with excess carbon produced 18.9 g Fe . Fe 2 O 3 + 3 C ⟶ 2 Fe + 3 CO
lapo4ka [179]

Answer:

1. 0.338 moles of Fe

2. 0.700 moles of Fe

3. 48.3%

Explanation:

This is the reaction:

Fe₂O₃ + 3C →  2Fe + 3CO

We were told that we produce 18.9 g of Fe. Let's convert the mass to moles:

18.9 g . 1mol/ 55.85 g = 0.338 moles of Fe

Let's make a rule of three; ratio is 1:2.

1 mol of oxide can produce 2 moles of elemental iron

Then, 0.350 moles must produce (0.350 .2) / 1 = 0.700 moles of Fe

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(Yield produced /Theoretical Yield) . 100 = 48.3 %

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4 years ago
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Which type of bond shows two or more atoms sharing electrons?
andre [41]

Answer:

Covalent is the type of bond.

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3 years ago
Part A Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 12N2(g)+32H2(g)→NH3(g) Express your answ
inessss [21]

Answer:

\Delta S^{0} for the given reaction is -99.4 J/K

Explanation:

Balanced reaction: \frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)

\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]

where S^{0} represents standard entropy.

Plug in all the standard entropy values from available literature in the above equation:

\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K

So, \Delta S^{0} for the given reaction is -99.4 J/K

7 0
4 years ago
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