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Mars2501 [29]
3 years ago
7

Is there a way to tell whether or not to put two double bonds or not. For example SeS2 seems to work if it has two double bonds

or a resonance structure.
Chemistry
1 answer:
Nina [5.8K]3 years ago
8 0
Your first step should be to analyse the compound. For example, if the compound is carbon, you know it always has a valence of four, so, if it has a formula C2H4 (ethylene) it obviously has a double bond. There are difficulties here because benzene C6H6 can be considered to have 6 1.5 C-C bonds, being aromatic.

A second step is to look at its structure. Double bonds are traditionally shorter than single bonds; triple bonds shorter still. Covalent bonds do have typical lengths, nevertheless you can still have problems.

<span>A third step is to consider reactivity. For example, if you have a C=C double bond, you can add, say, bromine to it Thus C2H4 gives C2H4Br2, and by adding two bromine atoms you know you have one double bond. Again, benzene becomes an awkward molecule, but because of this, you know benzene does not have double bonds in the traditional sense</span>
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UkoKoshka [18]
<h2>9. Given : P + O2 = P2O5</h2>

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P = (1) 4 | P = (2) 4

O = (2) 10 | O = (5) 10

Reaction Information

Phosphorus + Dioxygen = <u>Phosphorus Pentoxide</u>

Reaction Type: <u>Synthesis</u>

<h2>10. Given: HCl + NaOH = NaCl + H2O </h2>

<u>Reactant Product</u>

H = 2 | H = 2

Cl = 1 | Cl = 1

Na = 1 | Na = 1

O = 1 | O = 1

<h3>Chemical Equation is already in balance.</h3>

Reaction Information<em>:</em><em> </em><em>Hydrogen Chloride </em>+ <em>Sodium Hydroxide</em> = <em>Sodium Chloride </em>+ <em>Water</em>

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<u>How to balance chemical equation?</u>

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5 0
4 years ago
Suppose 10.0 mL of 2.00 MNaOH is added to (a) 0.780 L of pure water and (b) 0.780 L of a buffer solution that is 0.682 Min butan
katrin2010 [14]

Answer:

a) pH will be 12.398

b) pH will be 4.82.

Explanation:

a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles

The total volume after addition of pure water = 0.780+0.01 = 0.79 L

The new concentration of /NaOH will be:

molarity=\frac{molesofsolute}{volumeofsolution}=\frac{0.02}{0.79}=0.025M

the [OH⁻] = 0.025

pOH = -log [OH⁻] = 1.602

pH = 14 -pOH = 12.398

b) The buffer has butanoic acid and butanoate ion.

i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:

pH=pKa+log\frac{[salt]}{[acid]}

pKa=-logKa=-log(1.5X10^{-5})=4.82

ii) on addition of base the pH will increase.

8 0
4 years ago
A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated hydrogen perox
Nuetrik [128]

Answer:

2H_{2}O_{2} ---> 2H_{2}O + O_{2}

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These type of question rely on the person answering the question knowing the formula of hydrogen peroxide. Hydrogen peroxide has a formula that is very similar to that of water except it has a two after the O, which makes it very easy to remember.

6 0
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