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seraphim [82]
3 years ago
5

A 25.0 ml sample of an unknown monoprotic acid was titrated with 0.12 m naoh. the student added 31.6 ml of naoh and went past th

e equivalence point. which procedure could be performed next to more accurately determine the concentration of the unknown acid?
Chemistry
1 answer:
Inessa05 [86]3 years ago
7 0
The procedure, which can be used to determine more accurately the concentration of the unknown acid is TO BACK-TITRATE WITH ADDITIONAL HYDROCHLORIC ACID TO NEUTRALIZE THE ADDITIONAL SODIUM HYDROXIDE THAT WAS ADDED.
Monoprotic acids are acids that can donate only one proton per each molecule and they have only one equivalence point. Examples of monoprotic acids are HCI, HNO3 and CH3COOH.
The back titration method is typically used when one needs to determine the concentration of an analyte provided there is a known molar concentration of excess reactants. 
From the information given in the question above, we are told that excess NaOH was added. To correct this mistake, the right thing to do is to use additional HCl to carry out back titration, taking note of the quantity of acid that will be needed to neutralize the excess NaOH.
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Nataly [62]
Salutations!

Determine the length of the object shown below.

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5 0
3 years ago
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The lowest whole-number ratio of elements in a compound is called the ?
ASHA 777 [7]

Answer:

The Empirical Formula.

Explanation:

From the empirical formula and using the weight (in g)  of a given substance, we can come up with the molecular formula which is the actual weight of a substance. Sometimes, we find that the empircal formula is the molecular formula.

8 0
2 years ago
Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope
morpeh [17]

<u>Answer:</u>

<u>For a:</u> The isotopic representation of iodine is _{53}^{131}\textrm{I}

<u>For b:</u> The isotopic representation of cesium is _{55}^{137}\textrm{Cs}

<u>For c:</u> The isotopic representation of strontium is _{38}^{52}\textrm{Sr}

<u>Explanation:</u>

The isotopic representation of an atom is: _Z^A\textrm{X}

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

  • <u>For a:</u>

We are given:

Number of neutrons = 78

Atomic number of iodine = 53 = Number of protons

Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is _{53}^{131}\textrm{I}

  • <u>For b:</u>

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is _{55}^{137}\textrm{Cs}

  • <u>For c:</u>

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is _{38}^{52}\textrm{Sr}

3 0
3 years ago
Hurry please quick summary
Nadya [2.5K]
I think that work is being done on the books because they are being moved to their proper location and they will be sorted properly rather than lying on a table. Without lifting or carrying, you could sort the books by their genre or title name on the bookshelf so it will be sorted much more efficiently.

I’m not sure if this is the answer you are looking for but I hope it helps :)
3 0
3 years ago
If you have 3.45 × 1015 atoms of iron, how many moles of iron do you have?
kumpel [21]
3501.75

is the answer

i think

(your welcome)
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