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Georgia [21]
3 years ago
9

A sample of chlorine containing isotopes of mass numbers 35 and 37 was analysed in a mass-spectrometer. How many peaks correspon

ding to Cl+2 were recorded? A 2 B 3 C 4 D 5 Ans: B
Chemistry
1 answer:
Zolol [24]3 years ago
3 0
Three peaks corresponding to Cl+2 will be recorded. The peaks are for isotope 35, both 35 and 37 and for isotope 37. Mass spectrometer has the ability to detect and separate isotopes, even those differing by a single atomic mass unit. When chlorine isotopes are analysed by mass spectrometer, either peak M or M+2 can be obtained. The intensity ratio in the isotope pattern depends on the natural abundance of the isotopes.
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Which particle often described as having very little or no mass since the mass number does not change when you add more and more
astra-53 [7]

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I wrote this answer to get points lol

Explanation:

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Rank the following aqueous solutions from highest to lowest freezing point: 0.1 m FeCl3, 0.30 m glucose (C6H12O6), 0.15 m CaCl2.
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Explanation: See images below for explanation

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The carbon-carbon double bond in ethene is ________ and ________ than the carbon-carbon triple bond in ethyne.
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weaker and longer

Explanation:

Since there are 3 bonds in ethyne in comparision with the 2 bonds of ethyne between carbon atoms, they are attracted more to each other → the bond gets shorter . And since there are one more bond that supports the union → the bond gets stronger

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3 years ago
Combustion analysis is performed on 0.50 g of a hydrocarbon, and 1.47 g of CO2 and 0.902 g of H2O are produced. What is the empi
JulijaS [17]

1. The empirical formula of the hydrocarbon is CH₃

2. The molecular formula of the hydrocarbon is C₂H₆

<h3>How to determine the mass of Carbon </h3>
  • Mass of CO₂ = 1.47 g
  • Molar mass of CO₂ = 44 g/mol
  • Molar of C = 12 g/mol
  • Mass of C =?

Mass of C = (12 / 44) × 1.47

Mass of C = 0.4 g

<h3>How to determine the mass of H</h3>
  • Mass of compound = 0.5 g
  • Mass of C = 0.4 g
  • Mass of H = ?

Mass of H = (mass of compound) – (mass of C)

Mass of H = 0.5 – 0.4

Mass of H =0.1 g

<h3>1. How to determine the empirical formula </h3>
  • C = 0.4 g
  • H = 0.1 g
  • Empirical formula =?

Divide by their molar mass

C = 0.4 / 12 = 0.03

H = 0.1 / 1 = 0.1

Divide by the smallest

C = 0.03 / 0.03 = 1

H = 0.1 / 0.03 = 3

Thus, the empirical formula of the compound is CH₃

<h3>2. How to determine the molecular formula</h3>
  • Empirical formula = CH₃
  • Molar mass = 30 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

Learn more about empirical formula:

brainly.com/question/24297883

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