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Reika [66]
3 years ago
8

Rank the following in order of increasingly positive hydrogen charge: a) HClO2, HCl, CH4, H2O, HClO

Chemistry
1 answer:
Lisa [10]3 years ago
8 0

Answer:

CH4<HClO2<HClO<H2O<HCl

Explanation:

In ranking the compounds on the basis of increasing positive charge of hydrogen, we must consider the relative oxidation numbers of the various elements in the compounds.

Methane contains a nonpolar covalent bond, HClO2 contains a chlorine atom in +3 oxidation state, HClO contains a chlorine atom in +1 oxidation state. The more the positive charge on the halogen atom the lesser the magnitude of positive charge on the hydrogen atom. Between water and HCl, chlorine is more electro negative hence HCl will contain a more positively charged hydrogen atom than water.

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What is the specific gravity of a concentrated sulfuric acid solution? A. 1.419 B. 1.000 C. 1.835 D. 1.263
GaryK [48]
Hello.

Your answer would be:

It’s about 1.84

So in your case

C. 1.835 Rounded

Plz mark me brainliest
5 0
4 years ago
Read 2 more answers
18.The average or small size stars become a _________ at the end of their life cycle.
wel
The answer is B, Neutron star
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3 years ago
When carbon is completely combusted in the presence of oxygen the only product produced is carbon dioxide as shown in the follow
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6 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
Consider the molecules H2O, H2S, H2Se, and H2Te. Which do you expect to have the highest boiling point, and why?
STatiana [176]

Answer:

a. H20,because it experiences hydrogen bonding.

5 0
3 years ago
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