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Vitek1552 [10]
3 years ago
10

Based on listed oxidation states which formulas would most likely result when sulfur and oxygen combine to form a compound?

Chemistry
1 answer:
Elanso [62]3 years ago
6 0

Answer: SO₂ (Sulphur(iv)oxide)

Explanation:

<em>Oxidation:</em> Oxidation is the process involving a loss in electron, it can also mean the addition of oxygen to a substance during chemical reaction. It can also mean removal of hydrogen from a substance during chemical reaction. it can also mean addition of electropositive element and removal of electronegative element from a substance during chemical  reaction.

<em>Oxidation Number:</em><em> The oxidation number of an element in any particular molecule or ion is defined as the electrical charge.</em>

Sulphur have variable oxidation states. And the oxidation number of surphur

is -2, -4 or -6, and the oxidation number of oxygen is -2.

Therefore when sulphur react with oxygen, the most likely results is sulphur(iv)oxide. (SO₂).

<em></em>

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During _____, bonds between monomers are broken by adding water.
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During _<span>A.   hydrolysis</span>, bonds between monomers are broken by adding water.
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Which pair of substances would most likely result in the production of a gas when reacting with an acid?
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Explanation:

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M+xHCl(aq)\rightarrow MCl_x(aq)+xH_2(g)

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4 years ago
Read 2 more answers
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

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(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

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