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NNADVOKAT [17]
3 years ago
10

How many grams of glucose are in 11.5 moles?

Chemistry
1 answer:
qaws [65]3 years ago
5 0

There are 2071.4662 grams of glucose in 11.5 moles.

Per 1 mole there are 180.15588 grams of glucose. 180.5588 x 11.5 =2076.4262

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Balance the following unbalanced redox reaction (assume acidic solution if necessary): Cr2O72- + Cl- → Cl2 + Cr3+ Indicate the c
sergejj [24]

Answer:

14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O

The coefficient that will be used for Cl₂ in this reaction is 3

Explanation:

We use the method of electron-ion to the balance.

We assume that the redox reaction is happening at acidic medium.

Cr₂O₇²⁻ + Cl⁻ → Cl₂ + Cr³⁺

Chloride is raising the oxidation state from -1 in the chloride, to 0 in the chloride dyatomic. This is the half reaction of oxidation

2Cl⁻ → Cl₂ + 2e⁻         Oxidation

In the dichromate anion, chromium acts with +6 in oxidation state, and we have 2 Cr, so the global charge of the element is +12. To change to Cr³⁺ it has release 3 electrons, but we have 2 Cr, so it finally released 6 e-. The oxidation state was decreased, so this is the reduction half reaction.

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O   Reduction

As we have 7 O in the product side, we add 7 water, to the opposite place. In order to balance the H (protons) we, add the amount of them, in the opposite side, again.

(2Cl⁻ → Cl₂ + 2e⁻) ₓ3        

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O)  ₓ1    

We multiply the half reactions, in order to remove the electrons and we sum, the equations:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6Cl⁻ → 3Cl₂ + 6e⁻ + 2Cr³⁺ + 7H₂O

Now, that we have the same amount of electrons, they can cancelled, so the balanced redox reaction is:

14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O

4 0
2 years ago
Read 2 more answers
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