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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

nadechworld.com:

0.1M

Explanation:

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

**Explanation:**

Learn more:

Alkali metal and halogens brainly.com/question/6324347

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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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**nadechworld.com:**

CuSO₄.H₂O = 1

CuSO₄.3H₂O = 3

CuSO₄.5H₂O = 5

CuSO₄7.H₂O = 7

CuSO₄.9H₂O = 9

**Explanation:**

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