The oxidation states of the facets Fe and Br needs to be figured out.Fe is a transition facet therefore is capable of multiple oxidation claims. an ionic compound is made with Fe as the cation and halide ion as anionBr⁻ which is a halogen has actually a charge of -1. In ions the charge of the ion is equal to the oxidation state of the ion. Br⁻ has oxidation state of -1. FeBr₂ is a compound therefore net charge is 0.the amount of the commodities of the oxidation says of the individual facets by the variety of atoms must include upto the net chargeoxidation number of Fe cation + (oxidation variety of Br⁻ x 2) = 0say the oxidation state of Fe is X and also oxidation number of Br⁻ is -1X + (2 x -1)= 0X - 2 = 0X = 2this means that oxidation variety of Fe is +2and also oxidation variety of Br is -1

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In standardization of NaOH solution, a student discovered that 25.55cm3 of base neutralized precisely 21.35cm3 of 0.12M HCl. Find the m



We"ll start by composing a well balanced equation for the reaction between HCl and also NaOH. This is illustrated below:

HCl + NaOH —> NaCl + H2O

From the over equation, we obtained:

nA (mole of the acid) = 1

nB (mole of the base) = 1

File acquired from the question include:

Ma (Molarity of the acid ) = 0.12M

Va (volume of the acid) = 21.35cm3

Mb (Molarity of the base) =?

Vb (volume of the base) = 25.55cm3

Using the equation MaVa/MbVb = nA/nB, the molarity of the base can be obtained as follow:

MaVa/MbVb = nA/nB

0.12 x 21.35 / Mb x 25.55 = 1

Cross multiply to express in linear create.

Mb x 25.55 = 0.12 x 21.35

Divide both side by 25.55

Mb = (0.12 x 21.35)/25.55

Mb = 0.1M

Because of this, the molarity of the base (NaOH) is 0.1M


Group 1A contains the the majority of electropositive metal whereas Group VIIA consists of the a lot of electronegative non-steel. The valence shell of team 1A is consisted of of a single electron yet those of VIIA has salso electrons. Based on the valence electrons, Group 1A will certainly easily lose an electron in order to finish its octet via an oxidation state of +1Group VIIA are very reactive team of aspects through a dire should acquire an electron to complete their octet configuration. The most reactive steel is Francium whereas Fluorine is the most reactive non-metal.

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Alkali metal and halogens


C4H10 + 6.5 O2 ----> 4CO2 + 5H2O 2C4H10 + 13 O2 ----> 8CO2 + 10H2O 1. Count the C on the left (4), put a 4 wright here the C on the appropriate. 2. Count the H on the left (1), you have two on the right, so you multindicate this two by 5. Put the 5 in front of the H2O 3. Count the O on the right. You have 4*2 + 5 = 13. You have two on the left, so you require 6.5 on the left. 4. Now multiply every little thing on the equation by two so you have nice integer numbers. 5. examine you have actually the exact same amount of every little thing on each side. Example C: left 8, right 8, etc.I hope this helps. :)

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CuSO₄.H₂O = 1

CuSO₄.3H₂O = 3

CuSO₄.5H₂O = 5

CuSO₄7.H₂O = 7

CuSO₄.9H₂O = 9


Some salts as CuSO₄ are presented in the hydratated develop to offer some stcapacity in their offers.