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Rufina [12.5K]
3 years ago
14

What is the overall charge of the compound frbr

Chemistry
1 answer:
sammy [17]3 years ago
5 0

Answer:

Zero  

Explanation:

FrBr is an ionic compound .

Fr is in Group 1. Br is in Group 17.

The charges on the ions are +1 and -1, respectively.

The compound consists of Fr⁺Br⁻ ions.

However, there are equal numbers of + and - charges, so

The overall charge of the compound is zero.

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What is hydrogen ion concentration of an acid solution whose PH is 3.4?​
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3 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
How many moles of agcl are contained in 244 ml of 0.135 magcl solution? the density of the solution is 1.22 g/ml?
spin [16.1K]
Since we are only asked for the number of moles, we don't need the information of density. The concentration is expressed in terms of 0.135 M AgCl or 0.135 moles of AgCl per liter solution. The solution is as follows:

Moles AgCl = Molarity * Volume
Moles AgCl = 0.135 mol/L * 244 mL * 1 L/1000 mL
<em>Moles AgCl = 0.03294 mol </em>
3 0
3 years ago
Read 2 more answers
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