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frosja888 [35]
3 years ago
10

Identify the highlighted secondary lymphatic tissue located on the lateral wall of the oropharynx just inferior to the soft pala

te
Chemistry
1 answer:
Katena32 [7]3 years ago
6 0

Answer:

The correct answer is - palatine tonsils .

Explanation:

Palatine tonsils or more normally tonsils are the pair of soft tissue masses of highlighted secondary lymphatic tissue present on the lateral wall of the oropharynx in the tonsilar fossa just below to the soft plate. It is the lymphatic tissue that is located to inferior to the soft plate.

Thus, the correct answer is - palatine tonsils.

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Consider 5.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.30 L and the temperature is increased to 4
lakkis [162]

Answer:

P₂ = 1.12 atm

Explanation:

To find the new pressure, you need to use the Combined Gas Law:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

In this equation, "P₁", "V₁", and "T₁" represent the initial pressure, volume, and temperature. "P₂", "V₂", and "T₂" represent the new pressure, volume, and temperature. Before plugging the values into the equation, you need to

(1) convert the pressure from mmHg to atm (760 mmHg = 1 atm)

(2) convert the temperatures from Celsius to Kelvin (°C + 273)

The final answer should have 3 sig figs like the given values.

P₁ = 365 mmHg / 760 = 0.480 atm           P₂ = ? atm

V₁ = 5.00 L                                                   V₂ = 2.30 L

T₁ = 20°C + 273 = 293 K                             T₂ = 40°C + 273 = 313 K

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}                                              <----- Combined Gas Law

\frac{(0.480 atm)(5.00 L)}{293 K}=\frac{P_2(2.30 L)}{313 K}                       <----- Insert values

0.00819=\frac{P_2(2.30 L)}{313 K}                                     <----- Simplify left side

2.56 = P_2(2.30L)                                      <----- Multiply both sides by 313

1.12 = P_2                                                  <----- Divide both sides by 2.30

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What is the uncertainty of 300.0ft
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300.0+\-0.1

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