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lina2011 [118]
3 years ago
10

Match the element or group to the rule assigning its oxidation state.

Chemistry
2 answers:
slamgirl [31]3 years ago
6 0
It would go B. A. E. D. C.
Hope I helped!
Anna11 [10]3 years ago
5 0

C. Lone elements and atoms in gases : 0

Elements in the free elemental have an oxidation number of zero

E. Elements in groups 1, 2, and 17 and polyatomic ions : Ionic charge

Group 1, 2, and 17 ions are formed by the alkali metals, alkaline metals and halogens respectively. Alkali metals, alkaline metals and halogens form +1, +2 and -1  ions respectively. Polyatiomic ions have -1, -2, -3 charges such as OH-, CO32-, PO43- respectively.

B. Hydrogen : +1, – 1 if bonded to a diatomic metal

H has an oxidation number of +1. H in metal hydrides such as NaH (sodium hydride) has an oxidation number of -1

A. Oxygen : Almost always –2  

O has an oxidation number of -2.

D. Elements with multiple oxidation states : Determined by other elements in the compound

V (vanadium) has different oxidation which depends on the compound in which V is found.

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A 40.0-mL sample of 0.100 M HNO2 (Ka = 4.6 x 10-4 .) is titrated with 0.200 M KOH. Calculate: a. the pH when no base is added b.
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4 0
3 years ago
A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0
3 years ago
Help me please, will give brainly
geniusboy [140]

I don’t know sorry :(
6 0
3 years ago
Read 2 more answers
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