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Allisa [31]
3 years ago
12

Which is an important step in the alternate method for balancing equations in redox reactions?

Chemistry
2 answers:
Gennadij [26K]3 years ago
7 0

Answer:

C....  - Determining the half reactions of chemical equations

Explanation:

Try It... -

Whitepunk [10]3 years ago
6 0

Answer:

the answer is C

Explanation:

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Answer: Lung Cancer

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3 years ago
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Calculate ΔS° for the reaction: 4Cr(s) + 3O2(g) → 2Cr2O3(s), Substance: Cr(s) O2(g) Cr2O3(s), S°(J/K⋅mol): 23.77 205.138 81.2
Mariana [72]

<u>Answer:</u> The value of \Delta S^o for the reaction is 1051.93 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

4Cr(s)+3O_2(g)\rightarrow 2Cr_2O_3(s)

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cr_2O_3(s))})]-[(4\times \Delta S^o_{(Cr(s))})+(3\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(Cr_2O_3(s))}=881.2J/K.mol\\\Delta S^o_{(O_2(g))}=205.13J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (881.2))]-[(4\times (23.77))+(3\times (205.13))]\\\\\Delta S^o_{rxn}=1051.93J/K

Hence, the value of \Delta S^o for the reaction is 1051.93 J/K

5 0
4 years ago
Fill in the blanks with the correct words to complete the sentence.
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Answer:

Explanation:

Salt water intrusion can cause the <u><em>fresh</em></u> water in wells to become contaminated with<u><em> salt</em></u>water.

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Why are the elements fluorine (F) , chlorine (Cl) , and bromine (Br) all found in the same group within the periodic table? A) T
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Halogens:
- all have 7 electrons in their outer shell,
- are very reactive elements,
- form - 1 ions.
Answer: C ) They have same chemical properties. 
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4 years ago
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How many ml of naoh was used to reach the equivalence point?
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25 ml was used to reach the pint
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