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3 years ago

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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g Five calcite, CaCO3 (MW 100.085 g/mol), samples of equal mass have a total mass of 12.3±0.1 g. What is the absolute uncertaint

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Answer:

The value is $L = 0.985 \pm 0.00801 \ g$

Explanation:

From the question we are told that

The molar mass of $CaCO_3$ is $MW = 100.085 \ g/mol$

The total mass is $m_g = 12.3 \ g$

The uncertainty of the total mass is $\Delta g = 0.1$

Generally the molar weight of calcium is $M_c = 40 g/mol$

The percentage of calcium in calcite is mathematically represented as

$C = \frac{40.07}{100.085} * 100$

$C = 40.03 \%$

Generally the mass of each sample is mathematically represented as

$m= \frac{m_g}{5}$

$m= \frac{12.3}{5}$

$m= 2.46 \ g$

Generally mass of calcium present in a single sample is mathematically represented as

$m_c = 2.46 * \frac{40.04}{100}$

$m_c = 0.985 \ g$

The uncertainty of mass of a single sample is mathematically represented as

$k = \frac{\Delta g }{5}$

$k = \frac{0.1 }{5}$

$k = 0.02\ g$

The uncertainty of mass of calcium in a single sample is mathematically represent

$G = \frac{0.02 * 40.04}{ 100}$

$G = 0.00801 \ g$

Generally the average mass of calcium in each sample is

$L = 0.985 \pm 0.00801$

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