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3 years ago

Wuy Piolcululugu Give the names of two organs in the chest. 1. ............... 2 m

Chemistry

1 answer:

Arada [10]3 years ago

8 0

Answer:

The organs present inside the chest are :

1. The lungs

2. The heart

Explanation:

The chest cavity is also called as the thoracic cavity. It is the second largest hollow space of the body.In the bottom , it is enclosed by the diaphragm.

This cavity actually contain three space each round with mesothelium , pleural cavity and precardial cavity.

This contain the lungs , the tracheobronchial tree , the heart , the blood vessels which transport the blood between the heart and the lungs.

It also contain the esophagus .

Esophagus is the path through which the food passes from the mouth to the stomach.

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How do seismologists, or scientists who study earthquakes, know that the largest known earthquake to hit the region occurred aro

denis23 [38]

Answer:

By looking at the amount of time between the P and S wave on a seismographs recorded on a seismograph, scientists can tell how far away the earthquake was from that location. However, they can't tell in what direction from the seismograph the earthquake was, only how far away it was.

Explanation:

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3 years ago

A 17.9 mL sample of a 0.445 M aqueous hypochlorous acid solution is titrated with a 0.374 M aqueous sodium hydroxide solution. W

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Answer:

the pH at the start of the titration is 3

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3 years ago

All matter can be classified by their physical properties. Which substance can be classified as being soluble in water

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3 years ago

A client takes 30 ml of magnesium hydroxide and aluminum hydroxide with simethicone P.O. 1 hour and 3 hours after each meal and

labwork [276]

Answer:

Due to the short term of its action it has in the stomach environment

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3 years ago

The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is ________.

Naddika [18.5K]

Answer:

12.6.

Explanation:

We should calculate the no. of millimoles of KOH and HCl:

no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.

no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.

It is clear that the no. of millimoles of KOH is higher than that of HCl:

So,

[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.395 M) = 1.4.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.4 = 12.6.

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3 years ago