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Elena-2011 [213]
4 years ago
11

Every oxygen atom has 8 electrons. what can you learn from the electron configuration of oxygen

Chemistry
1 answer:
-BARSIC- [3]4 years ago
7 0
What you can learn, is since there are 2 in the first ring of electrons, that is full, the other 6 are valence electrons which can bond with others, and want to bond with them to finish the ring of 8 electrons.
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Sodium bicarbonate is reacted with concentrated hydrochloric acid at 37.0 °c and 1.00 atm. The reaction of 6.00 kg of bicarbonat
aleksklad [387]

Answer:

1837.89 Lt

Explanation:

The chemical reaction for this situation is:

NaHCO₃ + HCl → NaCL + H₂O + CO₂ ₍g₎

Where the mola mass we need are:

M NaHCO₃ = 84 g/mol

M CO₂ = 44 g/mol

As we have 6.00 Kg of sodium bicarbonate, then:

6 Kg NaHCO₃ = 71.43 moles of NaHCO₃

Due the stoichiometry of this chemaicl reaction:

1 mol NaHCO₃ = 1 mol CO₂

71.43 moles NaHCO₃ = 71.43 moles CO₂

And considering that CO₂ is an ideal gas, we can use the following formula:

PV=nRT

V = (nRT)/P

n = 71.43 mol

R = 0.083 Ltxatm(molxK)

T = 37°C = 310 K

P = 1 atm

So: V = (71.43x0.083x310)/1

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Answer:

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Balance the redox reaction Al(s) + MnO4^- (aq) --> MnO2 (s) + Al(OH)4^- (aq) in aqueous basic solution
GREYUIT [131]

Answer:

Al + MnO4- + 2H2O → Al(OH)4- + MnO2

Explanation:

First of all, we out down the skeleton equation;

Al + MnO4- → MnO2 + Al(OH)4-

Secondly, we write the oxidation and reduction equation in basic medium;

Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-

Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-

Thirdly, we add the two half reactions together to obtain:

Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-

Lastly, cancel out species that occur on both sides of the reaction equation;

Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O

The simplified equation now becomes;

Al + MnO4- + 2H2O → Al(OH)4- + MnO2

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