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il63 [147K]
3 years ago
15

Ned help with this question​

Chemistry
1 answer:
Liula [17]3 years ago
7 0
100/0.25 = 400 cm^3
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When a living organism is successful in meeting all its needs ,this is called
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homeostasis

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In biology, homeostasis (or “health”) is the state of steady internal, physical, and chemical conditions maintained by living systems. This is the condition of optimal functioning for the organism and includes many variables, such as body temperature and fluid balance, being kept within certain pre-set limits (homeostatic range). Other variables include the pH of extracellular fluid, the concentrations of sodium, potassium and calcium ions, as well as that of the blood sugar level, and these need to be regulated despite changes in the environment, diet, or level of activity. Each of these variables is controlled by one or more regulators or homeostatic mechanisms, which together maintain life.

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Match each of the follow two either Meiosis or Mitosis
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meiosis

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An oily liquid with a cheesy, waxy odor like that of goats or other barnyard animals is caproic acid, a compound containing carb
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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

Learn more about empirical formula:

brainly.com/question/9459553

#SPJ1

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b

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