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lora16 [44]
3 years ago
8

The formula in my book for carbonate is Co32- and I’m confused because carbon has 4 valence electrons and oxygen has 6 . How doe

s it end up being CO32-?? Please help
Chemistry
2 answers:
polet [3.4K]3 years ago
6 0
The total charge of the carbonate ion is -2. O2 in CO32- has a total charge of -6 (3 x -2). Hence, to find the charge of carbon -> x + -6 = -2. Therefore, carbon has a charge of +4.
This matches your theory stated whereby
1. Carbon has 4 valence electrons so in this case, it loses 4 electrons to obtain a stable noble octet config. & hence has a charge of +4.
2. Oxygen has 6 valence electrons so in this case, it gains 2 electrons & hence has a charge of -2.
balandron [24]3 years ago
4 0
Hello!!!

what exactly are you wanting to know? It isn’t clear from your question. Please advise and I will answer it for you ASAP. I think I know what you are wanting but I don’t want to assume.... thank you
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In the laboratory, a student dilutes 13.5 mL of a 11.6 M hydroiodic acid solution to a total volume of 100.0 mL. What is the con
igomit [66]

Answer: The concentration of the diluted solution is 1.566M.

Explanation:

The dilution equation is presented as this: M_{s} V_{s} =M_{d} V_{d}.

·M= molarity (labeled as M)

·V= volume (labeled as L)

·s= stock solution (what you started with)

·d= diluted solution (what you have after)

Now that we know what each part of the formula symbolizes, we can plug in our data.

11.6M*13.5mL=M_{d} *100mL

We cannot leave it like this because the volumes must be in Liters, not milliliters. To convert this, we divide the milliliters by 1000.

13.5mL/1000=0.0135L     100mL/1000=0.1L

Now that we have the conversions, let's plug them into the equation.

11.6M*0.0135L=M_{d} *0.1L

The only thing that we need to do now is actually solving the answer.

M_{d} =\frac{11.6M*0.0135L}{0.1L}       M_{d} =1.566M

From the work shown above, the answer is 1.566M.

I hope this helps!! Pls mark brainliest :)

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So I lost my sense of taste yesterday.
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Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

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