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Lunna [17]
3 years ago
13

In which species does sulfur have the highest oxidation number? s8 (elemental form of sulfur) h2s so2 h2so3 k2so4?

Chemistry
1 answer:
Mashutka [201]3 years ago
8 0
There are rules to assigning conventional oxidation numbers to some elements. For those unspecified, you can solve them. This is how you solve it.

*S8. Since this is in elemental form, its oxidation number is assigned as 0.

*H2S. H is assigned with +1. Since the compound is neutral, the overall charge is 0. So,
2(+1) + x = 0
x = -2
The charge of S here is -2.

*SO₂. O is assigned with (-2). Using the same procedure,
x + 2(-2) = 0
x = +2
The charge of S here is +2.

*H₂SO₃.
2(+1) + x + 3(-2) = 0
x = +4.
The charge of S here is +4.

*K₂SO₄. K is assigned with +1. 
2(+1) + x + 4(-2) = 0
x = +6
The charge of S here is +6.

<em>The S with the highest oxidation number is the one in K₂SO₄.</em>

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Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have
Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

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2 years ago
August kekule described the various ring structures of the carbon compound benzene. What type of chemist would he be considered
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Us
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Answer:

24.9%

Explanation:

According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:

0.0927 × 100 = 9.27%

Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;

100% - 9.27% = 90.73%

THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles

N.B: mole = mass/molar mass

Given the Molar Mass

NaCl: 58.44 g/mol

H2O: 18.016 g/mol

For NaCl;

0.0927 = mass/58.44

mass = 0.0927 × 58.44

5.42g

For H2O;

9.073 = mass/18.016

mass = 9.073 × 18.016

= 16.35g

Total mass of solution = 16.35g + 5.42g = 21.77g

Mass percent of NaCl = mass of NaCl/total mass × 100

% mass of NaCl = 5.42g/21.77g × 100

= 0.249 × 100

= 24.9%

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