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irakobra [83]
3 years ago
15

State where the chemical potential energy released from a reaction is stored

Chemistry
1 answer:
Vlad [161]3 years ago
4 0
It is stored in the bonds between atoms.
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Refer to the two formulas for xylose.
Ludmilka [50]

Answer:

A is the molecular formula for xylose because shows the actual number of atoms in the compound: Formula B is the empirical formula for xylose because it shows the smallest whole-number ratio for the different atoms in the compound: Formula A is the molecular formula for xylose because shows the arrangement of atoms in the compound: Formula B is the structurab formula for xylose because it shows the smallest whole-number ratio for the different atoms in the compound: Formula A is the empirical formula for xylose because it shows the actual number of atoms in the compound: Formula B is the molecular formula for xylose because it shows the smallest whole-number ratio for the different atoms in the compound: Formula A is the structural formula for xylose because it shows the arrangement of atoms in the compound: Formula B is the empirical formula for xylose because it shows the smallest whole-number ratio for the different atoms in the compound.

5 0
2 years ago
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the emission of the lowest-energy pho
Sladkaya [172]
N=6 ——> n=1
hope this helps!
5 0
3 years ago
PLEASE HELP When I combine Sprite with a sour candy, it starts to bubble a lot. Is this a physical or chemical change?
maksim [4K]

Answer:

B. chemical

Explanation:

Chemical change cannot go back to its original form

4 0
2 years ago
Read 2 more answers
Identify the oxidizing agent and the reducing agent in the reaction. 8H +( aq) + Cr 2O 7 2–( aq) + 3SO 3 2–( aq) → 2Cr 3+( aq) +
S_A_V [24]

Answer:

SO₃²⁻ is the reducing agent and Cr₂O₇²⁻ is the oxidizing agent.

Explanation:

Oxidation reaction:

3SO₃²⁻ (aq) + 3H₂O (l) → 3SO₄²⁻ (aq) + 6H⁺ (aq) + 6e⁻                  

Reduction reaction:

Cr₂O₇²⁻ (aq) + 14H⁺ (aq) + 6e⁻ → 2Cr ³⁺ (aq) + 7H₂O (l)

Now, adding the oxidation and the reduction reactions we get the full net reaction:

Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺ (aq) → 2Cr ³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)

Since, the S in SO₃²⁻, present in the +4 oxidation state is oxidized to +6 oxidation state in SO₄²⁻, by the loss of 2e⁻.

<u>Therefore, SO₃²⁻ is the reducing agent. </u>

And, the Cr in Cr₂O₇²⁻, present in the +6 oxidation state is getting reduced to +3 oxidation state, Cr ³⁺, by the gain of 6e⁻.

<u>Therefore, Cr₂O₇²⁻ is the oxidizing agent.</u>

3 0
3 years ago
According to dalton's model, the atom is a solid sphere. What would the alpha particles do when they hit the floor gold foil if
tester [92]
The alpha particle would be attracted to the atom<span />
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3 years ago
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