Answer : The incorrect option is, The most of the mass of the atom comes from the electron cloud.
Explanation :
There are three basic particles of an atom which are neutrons, protons and electrons.
The nucleus which is present in the center of an atom that contains the neutrons and the protons. The protons are positively charged and neutrons has no charge.
The outer region of an atom contains the electrons and the electrons are negatively charged.
As per given options, the statement which is the most of the mass of the atom comes from the electron cloud is incorrect statement because the most of the mass comes from the nucleus in which protons and neutrons are present.
One of the most worrisome waste products of a nuclear reactor is plutonium 239 (239Pu). This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus (4He +... Q: One of the most worrisome waste products of a nuclear reactor is plutonium 239 (239Pu<span>).</span>
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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Answer:
not sure about it .I also need an answer
They have different number of Neutrons and protons, so their masses are different
Hope this helps!
