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babunello [35]
3 years ago
8

Is cyclopentane an isomer of pentane? Explain.

Chemistry
1 answer:
Georgia [21]3 years ago
8 0
Cyclopentane is NOT an isomer of pentane.

Isomers must have the same molecular formula.

The molecular formula of pentane is C₅H₁₂, while the molecular formula of cyclopentane is C₅H₁₀.

The two molecules cannot be isomers.
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Both the oxygen you need to breathe and ozone are gases made only of oxygen. So why can't you survive by breathing ozone?
marshall27 [118]

Answer/Explanation: Two atoms of oxygen form the basic oxygen molecule--the oxygen we breathe that is essential to life. The third oxygen atom can detach from the ozone molecule, and re-attach to molecules of other substances, thereby altering their chemical composition.

8 0
3 years ago
Determine the correct name for the compound MG3N2
SSSSS [86.1K]
Mg3N2 is Magnesium nitride
5 0
2 years ago
Calculate the percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl on the assumption that t
stealth61 [152]

Answer:

1.089%

Explanation:

From;

ν =1/2πc(k/meff)^1/2

Where;

ν = wave number

meff = reduced mass or effective mass

k = force constant

c= speed of light

Let

ν =1/2πc (k/meff)^1/2  vibrational wave number for 23Na35 Cl

ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl

The between the two is obtained from;

ν' - ν /ν  = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2

Therefore;

ν' - ν /ν = [meff/m'eff]^1/2 - 1

Substituting values, we have;

ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2  -1

ν' - ν /ν = -0.01089

percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;

ν' - ν /ν * 100

|(-0.01089)|  × 100 = 1.089%

4 0
3 years ago
The empirical formula of an organic compound is C2H4O. The molecular mass of the compound is 176g/mol.
Brrunno [24]

Answer:

The molecular formula of the compound is C_{8}H_{16}O_{4}. The molecular formula is obtained by the following expression shown below

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is  C_{2}H_{4}O

Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.

Empirical formula mass of the compound = \left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}

n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Molecular formula = 4 \times C_{2}H_{4}O

Molecular formula is C_{8}H_{16}O_{4}

6 0
3 years ago
What does the number of valence electrons determine?
N76 [4]

Valence electrons are the electrons in the outermost shell of an element on the periodic table. Atoms want to be able to have a full outer shell and they can share or trade electrons in order to achieve this. Valence electrons are also super super important in chemical reactions. The number of valence electrons determines what group that specific atom or element is in on the periodic table. This affects the reactivity of the element.

4 0
3 years ago
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