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3 years ago

9

In a titration, 10.0 ml of 0.0750 M HCl(aq) is exactly neutralized by 30.0 ml of KOH(aq) of unknown concentration. What is the c

oncentration of the KOH(aq)solution?

1 answer:

Sphinxa [80]3 years ago

4 0

Answer: 0.0250

Explanation: 10 X 0.0750 = .75

.75 / 30 = 0.0250 M

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Read 2 more answers

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<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 $_{38}^{52}\textrm{Sr}$

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