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avanturin [10]
1 year ago
14

Which of the following regions does NOT match its description?

Chemistry
1 answer:
Fantom [35]1 year ago
8 0

The region that does not match its description is the fourth answer choice which is Abdominal region: spine

  • For the first answer choice - Cranial region: head

The cranial region encompasses the upper part of the head.

∴ The first answer choice matches its description.

  • For the second answer choice - Axillary region: armpits

The axillary region is an anatomical region under the shoulder joint where the arm connects to the shoulder. Therefore, it encompasses the armpits.

∴ The second answer choice matches its description.

  • For the third answer choice - Thoracic region: chest

The thoracic region runs from the base of the neck down to the abdomen. Therefore, it encompasses the chest

∴ The third answer choice matches its description

  • For the fourth answer choice - Abdominal region: spine

The abdominal region is divided into four quadrants which include are

1. Right upper quadrant fossa (RUQ)

2. Right lower quadrant fossa (RLQ)

3. Left lower quadrant fossa (LLQ)

4. Left upper quadrant fossa (LUQ)

It is also divided into nine (9) areas, all of which does not include the spine.

∴ The fourth answer choice does NOT match its description.

Hence, the region that does not match its description is the fourth answer choice which is Abdominal region: spine

Learn more here: brainly.com/question/11504303

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What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua
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Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Na_2S

Given mass = 15.5 g

Molar mass of Na_2S = 78.0452 g/mol

<u>Moles of Na_2S = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>

Given: For CuSO_4

Given mass = 12.1 g

Molar mass of CuSO_4 = 159.609 g/mol

<u>Moles of CuSO_4 = 12.1 g / 159.609 g/mol = 0.0758 moles</u>

According to the given reaction:

Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS

1 mole of Na_2S react with 1 mole of CuSO_4

0.1986 mole of Na_2S react with 0.1986 mole of CuSO_4

Available moles of CuSO_4 = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, CuSO_4 is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of CuSO_4 gives 1 mole of CuS

0.0758 mole of CuSO_4 gives 0.0758 mole of CuS

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

<u>Option B is correct.</u>

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3 years ago
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