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dangina [55]
3 years ago
7

The diagram shown represents cells seen in the low-power field of view of a compound light microscope. (1mm = 1000μm) Length of

one of these cells is approximately:
A) 200 mm

B) 400 mm

C) 500 mm

D) 2,000 mm
Biology
1 answer:
Zolol [24]3 years ago
5 0

Answer:

Where is the picture

Explanation:

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The bones from an animal found at an archaeological dig have a C614 activity of 0.10 Bq per gram of carbon. The half-life of C61
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C14 is an isotope used in radiocarbon dating techniques to date organic matter remains. The age of these bones is approximately<u> 6890 years.</u>

<h3>What is Carbon 14?</h3>

Carbon 14, also known as radiocarbon, is a radioactive carbon isotope.

Isotopes are the atoms of the same element -carbon- that vary in neutrons and, hence, in their massic number. They are alternative forms of the same element.

The radioactive C14 nucleus contains 6 protons and 8 neutrons and has a half-life of 5730 years.

The term half-life is a reference. It means that an organism that has been dead for 5730 years has half the C14 amount or concentration than the same organism had when it was alive.

Knowing the half-life of an element is useful to determine the age of the dead matter.

C14 is used in radiocarbon dating techniques or methods to estimate the age of fossils. This is a reliable technique used for dating organic samples that are less than 50,000 years old.

<u>Available data</u>:

  • The half-life of C14 is 5730 years
  • Bones activity of 0.10 Bq per gram of carbon

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

  • C14 T₀ ⇒ Amount of carbon in a living body. We know, by bibliography, that living organism activity is 0.23 Bq per gram of carbon. So, C14 T₀ = 0.23 Bq/g
  • C14T₁ ⇒ Amount of carbon in the dead body. C14T₁ = 0.1 Bq/g
  • λ ⇒ radioactive decay constant = (Ln2)/T₀,₅
  • T₀,₅ ⇒ The half-life of carbon 14 = 5730 years
  • T₀ = Time when the organism was alive
  • T₁ = Age of bones

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5730

<u>λ = 0.0001209</u>

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (0.1/0.23) = Ln 0.4347 =<u> - 0.833</u>

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 0.833 = - 0.0001209 x T₁

T₁ = - 0.833 / - 0.0001209

T₁ =  6889.99 ≅ <u>6890 years</u>

The bones are approximately<u> 6890 years.</u>

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What are the five functions of the human skeleton?
olga2289 [7]

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Support, movement, protection, blood cell production, calcium storage and endocrine regulation

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Which is a characteristic of pseudoscience?
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Answer:

Resistance to change

Explanation:

Indicators/Characteristics of pseudoscience:

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  • Over-reliance on confirmation rather than refutation
  • <u>Lack of </u><u>openness </u><u>to testing by </u><u>other experts</u>
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Haase D, Germing U, Schanz J, et al. New insights into the prognostic impact of the karyotype in MDS and correlation with subtyp
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We have produced a huge, novel data set that incorporates morphologic, clinical, cytogenetic, and follow-up information from 2124 patients with myelodysplastic conditions (MDSs) at 4 organizations in Austria and 4 in Germany. Cytogenetic examinations were effectively acted in 2072 (97.6%) patients, uncovering clonal anomalies in 1084 (52.3%) patients.

Numeric and underlying chromosomal irregularities were reported for every patient and partitioned further as indicated by the quantity of extra anomalies. Along these lines, 684 unique cytogenetic classes were distinguished. The effect of the karyotype on the normal flow of the illness was concentrated on in 1286 patients treated with steady consideration as it were.

Middle endurance was 53.4 months for patients with ordinary karyotypes (n = 612) and 8.7 months for those with complex inconsistencies (n = 166). A sum of 13 uncommon irregularities were related to great (+1/+1q, t(1q), t(7q), del(9q), del(12p), chromosome 15 inconsistencies, t(17q), monosomy 21, trisomy 21, and - X), middle (del(11q), chromosome 19 peculiarities), or poor (t(5q)) prognostic effect, separately.

The prognostic significance of extra anomalies differed significantly relying upon the chromosomes impacted. For all World Health Organization (WHO) and French-American-British (FAB) order framework subtypes, the karyotype gave extra prognostic data. Our examinations offer new experiences into the prognostic meaning of uncommon chromosomal irregularities and explicit karyotypic mixes in MDS.

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7 0
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