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Akimi4 [234]
3 years ago
10

Which part of the nervous system is a gray, wrinkly organ made up of millions of neurons? A. nerves B. brain C. spinal cord D. e

ye
wasn't exactly sure what category to put in
Chemistry
2 answers:
Mariulka [41]3 years ago
8 0
The correct answer is b brain







irakobra [83]3 years ago
6 0
The answer to your question is b the brain<span />
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How many liters of 1.2M solution can be prepared with 0.50 moles of C6H12O6
BigorU [14]
First you need to know the molecular weight of sugar (C6H12O6) which is 180.156g/mol
You have half a mole so you have 90.078g
If you wanted to make 1L of a 1.2M solution of glucose you would need 180.156*1.2=216.1872g
But you only have 90.078g
So you need to figure out how much this 90.078g will make if the solution must be 1.2M:
90.078g/216.1872g=xL/1L
solve for the X and you get 0.416666666...
so 416.7ml or 0.417L
5 0
3 years ago
Please help. Thank you:)
bearhunter [10]
These are guesses because there is no word bank or anything

A. Water
B. Housing
C. Plants
D. Animals
E. Warmth
5 0
3 years ago
Help me with please !!!
egoroff_w [7]

I would think the last one about the ozone layer
6 0
3 years ago
How to solve this: a small dog has consumed a dangerous amount of apap. the vet has ordered: rx acetylcysteine 150mg/kg/stat. yo
Fantom [35]
Given:
Stock dose/concentration of 20% Acetylcysteine (200 mg/mL)
150 mg/kg dose of Acetylcysteine
Weight of the dog is 13.2 lb

First we must convert 13.2 lb to kg:
13.2 lb/(2.2kg/lb) = 6 kg

Then we must calculate the dose:
(150 mg/kg)(6kg) = 900 mg

Lastly, we must calculate the dose in liquid form to be administered:
(900 mg)/(200 mg/mL) = 4.5 mL

Therefore, 4.5 mL of 20% Acetylcysteine should be given.
3 0
3 years ago
Part C: complete the third column <br> Part D: complete the fourth column
Helga [31]

Answer:

Part C: P2 = 0.30 atm

Part D: V1 = 16.22 L.

Explanation:

Part C:

Initial pressure (P1) = 2.67 atm

Initial volume (V1) = 5.54 mL

Final pressure (P2) =.?

Final volume (V2) = 49 mL

The final pressure (P2) can be obtained as follow:

P1V1 = P2V2

2.67 x 5.54 = P2 x 49

Divide both side by 49

P2 = (2.67 x 5.54)/49

P2 = 0.30 atm

Therefore, the final pressure (P2) is 0.30 atm

Part D:

Initial pressure (P1) = 348 Torr

Initial volume (V1) =?

Final pressure (P2) = 684 Torr

Final volume (V2) = 8.25 L

The initial volume (V1) can be obtained as follow:

P1V1 = P2V2

348 x V1 = 684 x 8.25

Divide both side by 348

V1 = (684 x 8.25)/348

V1 = 16.22 L

Therefore, the initial volume (V1) is 16.22 L

6 0
3 years ago
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