<span>Answer:
Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either.
Start with your 2 half reactions:
I- --> IO3-
Cr2O72- --> 2 Cr3+
Balance O by adding H2O:
I- + 3 H2O --> IO3-
Cr2O72- --> 2 Cr3+ + 7H2O
Balance H by adding H+:
I- + 3 H2O --> IO3- + 6 H+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O
Balance charge by adding e-:
I- + 3 H2O --> IO3- + 6 H+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O
Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give:
Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
<span><span>When you write down the electronic configuration of bromine and sodium, you get this
Na:
Br: </span></span>
<span><span />So here we the know the valence electrons for each;</span>
<span><span>Na: (2e)
Br: (7e, you don't count for the d orbitals)
Then, once you know this, you can deduce how many bonds each can do and you discover that bromine can do one bond since he has one electron missing in his p orbital, but that weirdly, since the s orbital of sodium is full and thus, should not make any bond.
However, it is possible for sodium to come in an excited state in wich he will have sent one of its electrons on an higher shell to have this valence configuration:</span></span>
<span><span /></span><span><span>
</span>where here now it has two lonely valence electrons, one on the s and the other on the p, so that it can do a total of two bonds.</span><span>That's why bromine and sodium can form </span>
<span>
</span>
<span>Scientists ignore the forces of attraction between particles in a gas under ordinary conditions</span><span> because the particles in a gas are apart and moving fast, rather than clustered and moving slow, therefore the forces of attraction are too weak to have a visible effect.</span>
Answer:
see explanation below
Explanation:
You are missing the reaction scheme, but in picture 1, I found a question very similar to this, and after look into some other pages, I found the same scheme reaction, so I'm gonna work on this one, to show you how to solve it. Hopefully it will be the one you are asking.
According to the reaction scheme, in the first step we have NaNH2/NH3(l). This reactant is used to substract the most acidic hydrogen in the alkine there. In this case, it will substract the hydrogen from the carbon in the triple bond leaving something like this:
R: cyclopentane
R - C ≡ C (-)
Now, in the second step, this new product will experiment a SN2 reaction, and will attack to the CH3 - I forming another alkine as follow:
R - C ≡ C - CH3
Finally in the last step, Na in NH3 are reactants to promvove the hydrogenation of alkines. In this case, it will undergo hydrogenation in the triple bond and will form an alkene:
R - CH = CH - CH3
In picture 2, you have the reaction and mechanism.
