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3 years ago

Chlorine and bromine are elements in Group VII.

Chemistry

1 answer:

Goryan [66]3 years ago

7 0

Explanation:

Bromine is a chemical element with the symbol Br and atomic number 35. It is the third-lightest halogen, and is a fuming red-brown liquid at room temperature that evaporates readily to form a similarly coloured vapour. Its properties are intermediate between those of chlorine and iodine. Isolated independently by two chemists, Carl Jacob Löwig (in 1825) and Antoine Jérôme Balard (in 1826), its name was derived from the Ancient Greek βρῶμος ("stench"), referring to its sharp and disagreeable smell.

Bromine, 35Br

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Concave is not a type of mechanical wave because it doesn’t need a medium for propagation.

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What is an example of a scope three carbon emission?

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Production of materials and transportation are the examples of three carbon emission.

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3 years ago

1. How many times more acidic is Lemon Juice than apples?

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3 years ago

Find the empirical formula of each of the following compounds. Given mass or for each element in a sample of the compound 3,611

KengaRu [80]

Answer:

CaCl₂

Step-by-step explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Ca to Cl.

Data:

Mass of Ca = 3.611 g

Mass of Cl = 6.389 g

Calculations

Step 1. Calculate the moles of each element

Moles of Ca = 3.611 g Ca × (1 mol Ca/(40.08 g Ca)= 0.090 10 mol Ca

Moles of Cl = 6.389 g Cl

Step 2. Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

Ca:Cl = 0.090 10:0.1802 = 1:2.000

Step 3. Round the molar ratios to the nearest integer

Ca:Cl = 1:2.000 ≈ 1:2

Step 4: Write the empirical formula

EF = CaCl₂

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3 years ago

Chromium has an atomic mass of 51.9961 u and consists of four isotopes, 50 Cr , 52 Cr , 53 Cr , and 54 Cr . The 52 Cr isotope ha

Troyanec [42]

Answer : The atomic mass of $_{24}^{53}\textrm{Cr}$ isotope is 52.8367 amu

Explanation :

We know that:

Total percentage abundance of the isotope = 100 %

Percentage abundance of $_{24}^{50}\text{Cr}\text{ and }_{24}^{53}\textrm{Cr}\text{ isotopes}=[100-(83.79+2.37)]=13.84\%$

We are given:

Ratio of $_{24}^{50}\textrm{Cr}\text{ and }_{24}^{53}\textrm{Cr}$ isotopes = 0.4579 : 1

Percentage abundance of $_{24}^{50}\textrm{Cr}$ isotope = $\frac{0.4579}{(0.4579+1)}\times 13.84\%=4.37\%$

Percentage abundance of $_{24}^{53}\textrm{Cr}$ isotope = $\frac{1}{(0.4579+1)}\times 13.84\%=9.49\%$

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows: $\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the mass of $_{24}^{53}\textrm{Cr}$ isotope be 'x'

For $_{24}^{50}\textrm{Cr}$ isotope:

Mass of $_{24}^{50}\textrm{Cr}$ isotope = 49.9460 amu

Percentage abundance of $_{24}^{50}\textrm{Cr}$ = 4.37 %

Fractional abundance of $_{24}^{50}\textrm{Cr}$ isotope = 0.0437

For $_{24}^{52}\textrm{Cr}$ isotope:

Mass of $_{24}^{52}\textrm{Cr}$ isotope = 51.9405 amu

Percentage abundance of $_{24}^{52}\textrm{Cr}$ isotope = 83.79 %

Fractional abundance of $_{24}^{52}\textrm{Cr}$ isotope = 0.8379

or $_{24}^{53}\textrm{Cr}$ isotope:

Mass of $_{24}^{53}\textrm{Cr}$ isotope = x amu

Percentage abundance of $_{24}^{53}\textrm{Cr}$ isotope = 9.49 %

Fractional abundance of $_{24}^{53}\textrm{Cr}$ isotope = 0.0949

For $_{24}^{54}\textrm{Cr}$ isotope:

Mass of $_{24}^{54}\textrm{Cr}$ isotope = 53.9389 amu

Percentage abundance of $_{24}^{54}\textrm{Cr}$ isotope = 2.37 %

Fractional abundance of $_{24}^{54}\textrm{Cr}$ isotope = 0.0237

Average atomic mass of chromium = 51.9961 amu

Putting values in equation 1, we get:

$51.9961=[(49.9460\times 0.0437)+(51.9405\times 0.8379)+(x\times 0.0949)+(53.9389\times 0.0237)]\\\\x=52.8367amu$

Hence, the atomic mass of $_{24}^{53}\textrm{Cr}$ isotope is 52.8367 amu

4 0

4 years ago