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Nady [450]
3 years ago
11

A patient can't move his legs after a motorcycle accident. The doctor suspects nerve damage. What part of the peripheral nervous

system could be injured? (3 points) Select one: a. Motor neuron b. Cerebrospinal fluid c. Dura mater d. Grey matter
Chemistry
2 answers:
Daniel [21]3 years ago
6 0
A motor neuron is the part of the peripheral nervous system that could be injured.
Nutka1998 [239]3 years ago
5 0
Motor neuron! it included volentoury movement.
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Use molecular structure and intermolecular bonding to describe why bromine has a lower boiling point than water
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Answer:

Bromine mollecules are held together by van der waals forces while a water molecule constitutes both van der waals forces and hydrogen bomnding

Explanation:

This makes the water molecule recquire more heat energy to break the bond thus a higher boiling point while bromine structure requires just litttle heat energy  

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Determine which is the control group and which is the experimental group in the following scenario ( Scientists are testing a ne
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If 30 grams of KCl is dissolved at 10°C, how many additional grams would be needed to make the solution saturated at 60°C? * Cap
MariettaO [177]

If 30 grams of KCl is dissolved at 10°C, 14 g of KCl should be added to make a saturated solution at 60 °C.

<h3>What is a saturated solution?</h3>

A saturated solution is a solution in which there is so much solute that if there was any more, it would not dissolve. Its concentration is the same as the solubility at that temperature.

  • Step 1. Calculate the mass of water.

At 10 °C, the solubility is 31.2 g KCl/100 g H₂O.

30 g KCl × 100 g H₂O/31.2 g KCl = 96 g H₂O

  • Step 2. Calculate the mass of KCl required to prepare a saturated solution at 60 °C.

At 60 °C, the solubility is 45.8 g KCl/100 g H₂O.

96 g H₂O × 45.8 g KCl/100 g H₂O = 44 g KCl

  • Step 3. Calculate the mass of KCl that must be added.

44 g - 30 g = 14 g

If 30 grams of KCl is dissolved at 10°C, 14 g of KCl should be added to make a saturated solution at 60 °C.

Learn more about saturated solutions here: brainly.com/question/24564260

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D. Anti-periplanar

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In the <u>second step</u> of the intramolecular William Ether Synthesis mechanism (figure 1) we will have the attack of the negative charge of the oxygen to the carbon bond to the Br. At the same time the Br leaves, so a bond would be broken (the <u>C-Br</u> bond) and a bond would be formed (the <u>C-O</u> bond).

Now, this process can happen only if the <u>attack</u> and the <u>leaving group </u>has an anti configuration (figure 2). In an anti configuration the <u>nucleophile</u> and the <u>leaving group</u> would have <u>opposite directions</u>.

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