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maxonik [38]
2 years ago
15

The chemical formula of bromomethane is BrCH3. Which correctly describes the number of atoms of each element in bromomethane?

Chemistry
1 answer:
Anit [1.1K]2 years ago
4 0

The number of atoms of each element :

C : 1 atom

H : 3 atoms

Br = 1 atom

<h3>Further explanation</h3>

Given

Bromomethane-CH₃Br

Required

The number of atoms

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.  

A molecular formula is a formula that shows the number of atomic elements that make up a compound.  

The number of atoms in a compound is generally indicated as a subscript after the atom

C : 1 atom

H : 3 atoms

Br = 1 atom

Total 5 atoms

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21 Given the following acid dissociation constants, which acid has the strongest conjugate base?
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When the pressure that a gas exerts on a sealed container changes from 1100 bar to 75.5 bar, the temperature changes from k to 2
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Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
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parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
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7 0
3 years ago
If a compound has a molar más of 180g/mol and it’s empirical formula is CH^2O, what is it’s molecular formula?
lesya [120]

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H: 1.00784 g/mol ≅ 1.008 g/mol

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4 0
2 years ago
Which element has a greater electronegativity?<br> fluorine (9) or radium (88)
emmainna [20.7K]

Answer:

Fluorine

General Formulas and Concepts:

<u>Chemistry</u>

  • Reading a Periodic Table
  • Periodic Trends
  • Electronegativity - the tendency for an element to attract an electron to itself
  • Z-effective and Coulomb's Law, Forces of Attraction

Explanation:

The Periodic Trend for Electronegativity is up and to the right of the Periodic Table.

Fluorine is Element 9 and has 9 protons. Radium is Element 88 and has 88 protons. Therefore, Radium has a bigger Zeff than Flourine.

However, since Radium is in Period 7 while Fluorine is in Period 2, Radium has more core e⁻ than Fluorine does. This will create a much larger shielding effect, causing Radium's outermost e⁻ to have less FOA between them. Fluorine, since it has less core e⁻, the FOA between the nucleus and outershell e⁻ will be much stronger.

Therefore, Fluorine would attract an electron more than Radium, thus bringing us to the conclusion that Fluorine has a higher electronegativity.

7 0
2 years ago
Read 2 more answers
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